
















THB 



ILLUSTRflTBD 

Comprising an original treatise on Addition, Subtraction, 
Multiplication ana Division, showing how these processes 
may be shortened when applied to the practical com¬ 
putations of business, together with easy, short 
methods in Fractions, Decimals, Interest, and 
many rapid rules in buying and selling the 
various articles of commerce, to which is 
added rules for finding the day of the 
week corresponding to any date 
forever, Easter Sunday, etc., 

TOGETHER WITH 

A large but ingeniously condensed collection of 
simple rules for all kinds of 
measurements. 

THE WHOLE DESIGNED AS 

A SELF INSTRUCTOR 

For Home study, a Guide and Reference for the accountant and practical man 
in the office and workshop, and as a text-book for special 
classes in Schools and Business Colleges. 

BY 

WM. K DAYID. 

1902 . 

WM. K. DAVID, PUBLISHER, 

829 CAMBRIA STREET, COR. NINTH, 
PHILADELPHIA. 

Sent postpaid on receipt of price, $1.00* 







COPYRIGHT, 1889, BY WM. K. DAVID. 
COPYRIGHT, 1896, BY WM. K. DAVID. 
ALL RIGHTS RESERVED. 


Comprising an original treatise on Addition, 
Subtraction, Multiplication, and Division, 
showing how these processes may be greatly 
shortened and applied to practical computa¬ 
tions, together with easy short methods in 
Fraction^, Decimals, Interest, and many rapid 
rules in buying and selling the various articles 
of commerce; to which is added rules for find¬ 
ing the day of the week corresponding to any date forever, Easter 
Sunday, etc., together with a large but ingeniously condensed collec¬ 
tion of simple rules for all kinds of measurements met with in practical 
life. The whole designed as a Self Instructor for home study, a guide 
and reference for the accountant and practical man in the office and 
workshop, and as a text-book for special classes in schools and business 
colleges. By Wm. K. David. 

Illustrated, and Elegantly Bound in Cloth. Price, postpaid, $1.00. 

Embracing an unusual collection of money¬ 
making, money-saving and health-giving pre¬ 
scriptions, receipts, formulas, processes and 
trade secrets. Secured at considerable expense 
from a multitude of thinkers and workers in 
practical affairs, and edited by Wm. K. David. 

Illustrated, and Elegantly Bound in Cloth. 
Price, postpaid, $1.25. 

A beautiful metal pendant for watch chain or 
pocket. The correct calendar for over 100 years 
can be instantly adjusted to any date. It con¬ 
sists of an aluminum wheel (size of silver 
quarter) turning upon a gold plate body in odd 
design. Sent together with card calendar, of 
same design, for hanging on wall, containing 
directions. Price, postpaid, $1. 00. 

A MIRROR CALENDAR. Consists of David’s 
Ideal Perpetual Calendar, and a fine French 
plate mirror, size 6x8, together, incased in the 
most beautiful frame ever devised. Celluloid 
imitations of ebony, opal, ivory, onyx, or pink 
and blue silk, gold plate corners, with sub¬ 
stantial nickel holder for setting on desk or hanging on wall. The 
calendar shows through an opening cut in the face, and the wheels are 
turned from the back to set it for any month in any year. A mirror 
and calendar appropriately combined. Beautiful present for lady or 
gentleman. Cannot be sent by mail. Sent by express to any part of 
United States, on receipt of.$2.25. 

LIVE AGENTS WANTED. 

ADDRESS ALL ORDERS TO... 

Wm. K. David, 

PUBLISHER AND MANUFACTURER, 

No. 829 Cambria St., Cor. Ninth, - PHILADELPHIA, PA« 

5 1 , ■> 

V' 

The JOHN J. and HANNA M. McMANUg 
and MORRIS N. and CHESLEY V. YOUNG 
Collect-on 

!u,. Gift—Oct. 12, 195& 



WATCH CHARM 

PERPETUAL 

CALENDAR 

2 UNIQUE 4 
USEFUL 

NOVELTIES 

IN I 






4-0 A 

500 


PREFACE. 


This book is designed for thinkers, young and old, who 
have acquired the rudiments of numbers in any ordinary 
arithmetic. It arrangement is convenient for special class 
teaching in schools and business colleges, or to be used by 
teacher or student in connection with the numerous arith¬ 
metical text books. It is especially designed as a self¬ 
instructor and referee for the practical accountant and user 
of figures in every line of work in this age of quick thought 
and rivalry in business. The methods and forms are short, 
simple, easily comprehended, and correct. The majority 
of the rules and tables are original, and cannot be found in 
any other work. To those who are familiar with the con¬ 
tents of School Arithmetics, Ready Reckoners, Calculators, 
etc., this will be apparent at a glance. 


THE AUTHOR. 





f 


4 




a, 






















Pl/SIWESS ApplTloW. 



fiKOUPIlfG METHOD. 

To add qnickly and correctly is an accomplishment 
that can be attained by any person who will diligently follow 
our instructions. Thousands of pages have been written on 
the subject of rapid or liglitiling; addition. Some 
authors advocate methods of adding two or more columns at 


































14 


THE SHORT-RULE ARITHMETIC. 


once, bnt experience has proven that but few can afford to 
spend their time in long hours of practice necessary to master 
it; and others have taught what is commonly known as the 
word system, requiring many pages of explanation, which 
after all is simply a system of grouping. 

The author has fully demonstrated that by adding a 
single column at a time the student can learn to grasp the 
sum of at least three figures at a glance, and by uniting 
the result to the sum of other groups as they are seen, an ex¬ 
ample can be added with greater ease and speed than 
by any other system. 

This of course requires practice, not only in perceiving the 
sum of each group quickly, but in adding the groups. The 
first thing to learn is to name the sum of any three digits as 
quickly as to pronounce a syllable of three letters. We 
choose the combination of three figures in preference to 
combinations of twos, because by a little practice the sum 
of three figures can be seen in an instant, and a great saving 
of time is made in adding the groups, as there are fewer of 
them and they are just as easily added as smaller groups. 
The beauty of learning to group figures in this manner will 
become apparent after the student is thoroughly familiar 
with groups of threes. Frequently the eye will intuitively 
pass on beyond three figures, and in an instant—like pro¬ 
nouncing a long word—perceive the sum of six, seven, eight, 
nine, or more figures at a glance and at the expenditure, 
seemingly, of no mental effort. 

The following table is arranged for practice and drill, and 
shows all the possible combinations of the nine digits in 
groups of three. It is not necessary to memorize the groups, 
but familiarize yourself with them until you can easily read 
their sums. This can be done by glancing over the table at 
random, or by writing on paper or slate any number of 
columns three deep and adding them as quickly as possible. 
When this is acquired practice adding the combinations 
together by taking longer columns of figures: 




ADDITION . 


15 


GROUPING ADDITION TABLE 

Containing all combinations of the nine digits in groups 
of three. It will be observed that no group can add more 
than 27 nor less than 3: 






16 


THE SHORT-RULE ARITHMETIC. 


The following example will explain the process, together 
with an unique method of disposing of the carrying figures: 


Commencing at the bottom of the right-hand column 
we find the sum of the first group is 11, which, added to 
the sum of the next group (10),=21. Proceed in like 
manner with the remaining groups, adding each new 
group to the sum of the groups already found, until the 
complete sum (71) of the column is obtained. Write 
the 7 (carrying figure) over the next column and the unit 
figure (1) under the column added. It will be observed 
that we can now leave our work, and upon resuming 
our addition the carrying figure being in its proper place 
we add it to the sum of its column, which we find to 
be 83, as follows: 

10+16-f9+18-bl3+10-}- 7 =83. 

Write the 8 above the next column and 3 below the col¬ 
umn added. The sum of the next column, found as 
before (including carrying figure), is 95, which place 
in the answer and the sum is complete. 

To obviate the possibility of mistaking the carrying 
figures for an entry, a light line may be drawn above the 
columns and the carrying figures written above it slightly 
smaller than the others, in red ink or with a lead pencil. 
If the sum to carry contains two figures write them over 
the proper column as in the case of one figure. 


PRACTICAL PROOF OF ADDITION. 

Writing the carrying figures in this manner affords a neat 
and decidedly accurate proof of addition, as follows: Com¬ 
mence with the carrying figure of the left-hand column 
and add downward. If the result equals the figures found 
below the column in the answer (95) we can fairly conclude 
that thus far we are correct. Begin with the carrying figure 
of the next column and we find the sum is 83, which corre¬ 
sponds to the 8 over the preceding column and the 3 in the 
answer. The sum of the last column added downward is 71* 


9 13 
345 > 
852 , 
3 4 
9 40 
362 
753 
191 
244 
90 7 
505 - 
638 
7 70 
242 * 
45 2 
005 
524 * 
12 2 , 

95 31 











SUBTRACTION. 


17 


which corresponds to the 7 over the preceding column and 
the 1 in the answer. A very little practice will enable any 
one to add upward or downward with equal facility. 


SUpTRACTlOl^. 



SHORT BUSINESS METHODS. 

There are many cases where substraction may be more 
easily performed by a process of adding. In subtrac¬ 
tion of simple numbers it does not surpass the ordinary 
method to any great extent, although many accountants who 
are familiar with the process prefer it. In obviating the 
2 








































18 


THE SHORT-RULE ARITHMETIC. 


From 7 8 6 5 
Deduct 5 7 9 4 


tedious subtractions in long division (see division), and also 
in many special cases met with quite often in actual busi¬ 
ness, the principle can be applied to decided advantage. 

We will first explain the method as applied to simple 
numbers: 

Explanation.— We begin at the right, as in 
the ordinary rule. Our object is to find the figure 
which when added to the 4 will make it equal 
the figure above it. This figure is 1, which we *> 41 ^ t 

write in the answer. When the figure below is 4 

larger than the corresponding figure above we must find a figure 
which when added to the lower figure will make the unit figure 
of their sura equal the upper figure, as follows: We see that 9 is 
greater than the 6 above it, therefore we must make 9 equal 16. To do 
this requires a 7. Write the 7 in the answer and carry the 1 to the next 
lower figure (7) we get 8; 8 lacks O to make it equal the 8 above it, so 
write O in the answer. We next perceivethat it added to 5 will make 
it equal the 7 above it, and by writing the !t in the answer beneath we 
have the complete remainder. 

EXAMPLES FOR PRACTICE. 


From 9 8 7 2 
Deduct 6 5 81 


From 8 7 5 9 6 
Deduct 7 9 416 


From 6 5 2 8 3 
Deduct 4 7 16 4 


329 1 


8180 


18119 


BOOK-KEEPERS’ SUBTRACTION. 

Frequently several numbers or entries are to be taken 
from a given amount; as, for instance, a firm sells a bill of 
goods amounting to $827.56 on which there were several 
payments, as follows: $125.45, $236.50, $84.75, $72.25. How 
much remains unpaid? 


Ordinary method. 


125.45 
2 3 6.50 
8 4.75 
7 2.25 

5 18.95 


18 2 7.56 
5 18.95 

308.6 1 


Short Rule. 

$ 827.56 

12 5.45 
2 3 6.50 
84.75 
72.25 

$ 308.6 1 













SUBTRACTION . 


19 


Explanation. —The sum of the first right-hand column of payments 
is 15. The figure necessary to make the unit figure o*f this sum equal 
the corresponding figure (6) of the subtrahend is 1, i. e., 1 added to 
15=16; we write the 1 in the answer and carry 1 (tens) to the sum of the 
next column of payments, which gives 19. As the 9 in 19 is greater than 
the 5 in the subtrahend we find it will take 6 to make the unit figure 
equal it; 19 + 6=25, so we write the 6 in the answer and carry 2 to the 
sum of the next column of payments, which gives 19; 8 +19=27, making 
the unit figure equal the corresponding figure of the subtrahend. Writ¬ 
ing down the 8 and carrying 2 to the sum of the next column we get 22. 
22 requires O to make its unit figure equal the 2 above, so we write down 
O and carry 2 to the sum of the next column, which gives 5. We find 
that 3 added to 5 will equal the 8 of the subtrahend, which, when 
written below, completes the remainder. While this requires consid¬ 
erable trouble to explain properly, it will be found very easy in practice. 


DISCOUNT SUBTRACTION. 

This principle affords a decidedly neat and very easy short 
rule for deducting a required percentage from any number. 
For instance, a wholesale house sells a bill of goods amount¬ 
ing to $762.00 allowing a discount of 8# for cash. What is 
the net amount: 


Short Rule. 

$7 63.00 
.08 

$ 701.04 


Ordinary Method. 

$ 763.00 

.08 

$ 60.9600 

$ 763.00 
6 0.96 

$ 701.04 


Explanation.— We simply multiply 762 by 8 and write in the answer 
the figures which when added to the product will equal $762.00, as 
follows: 8X2=16. Requiring a 4 to make 20 (next higher order of tens, 
as the figure on the right is 0,) we write 4 in the answer and carry the 2. 
8X6=48, to which we add the 2, which gives 50. As its unit figure is 0, 
which corresponds to the 0 above, we writs down O in the answer and 
carry 5 to the product of 8X7, which gives 61. 1 added to 61 will equal 
the 62 above, so we write down 1 and a O under the 6, and bringing 
down the remaining 7 we get the complete remainder. When cents 
occur in the amount multiply your per cent by the cents and reserve the 









20 


THE SHORT-RULE ARITHMETIC. 


tens figure of your product as a carrying figure to add to your product of 
the per cent by the first dollar figure, as follows: 

We perceive that 6 (#) multiplied by the 5 From, $256.50 
(dimes figure) equals 30; disregarding the 0 
we carry 3 to the product of 6X6, which 
gives 39. Then, proceeding as above, we 
always get the answer to the nearest cent. 


Deduct 


6 % 


$241.11 


The preceding rule is preferable to any other where the 
per cent of discount can be represented by a single ilgure, 
but where the per cent is represented by two or more 
figures, as 30$, 45$, etc., it is best to proceed as follows: 


From a bill of $926.75 deduct 40$. What is the net bill? 


Explanation.— The ordinary method is to find 40# of the amount of 
the bill and then subtract the percentage thus found from it. We can 
obviate this long subtraction by taking the difference of tne rate per 
cent of discount and 100#, which can be easily done mentally, and mul¬ 
tiply this difference by the amount of the bill. In this example the 
discount is 40#, which lacks 60# of 100#, or 60 cents of each dollar re¬ 
mains to be paid. We therefore simply multiply the nun. per of dollars 
by .60. $926.75X.60=$556.05, answer. If the discount had been 30% we 
would have multiplied by .70, if 50# by .50, if 60% by .40, if 65# by .35, etc 

Frequently a series of rates per cent of discount are 
adopted by jobbers and manufacturers, as for instance, 30$, 
20 $, and 10 $. The object in this case is not to give a discount 
of 60 $ straight, but to first deduct 30 $ from the list price 
and 20$ from the remainder and 10$ from the remainder 
thus obtained which will give the net price of their wares. 

A jobber sells a bill of hardware amounting to $400, at a 
discount of 40$, 30$, and 10$. 

Explanation.— This may be performed in two ways quite different 
from the ordinary method and much easier. First, we can deduct 40% 
from 100#, which gives 60%; $400X.60=$240 (first remainder). We now 
deduct the next per cent in the series (30%) from 100#, which gives 70%; 
$240X.70=$168 (second remainder). The last per cent in the series being 
10 we deduct it from 100#, which gives 90%; $168X.90=$151.20, net 
amount of bill. Another method is to subtract each rate per cent of the 
series separately from 100#, and by multiplying all the remainders to¬ 
gether we get the net per cent to be paid. In the above example we 
would take the differences of the per cents and 100% and multiply them 
together, as follows: .60X.70X.90=.378 or 37&#of the list price to be 
paid; $400X.378=$151.20, answer same as above. 100#— 37&# =62^#, 
which is the net per cent discount of the series. The net discount of 




MULTIPLICATION. 


21 


any series can be found in this manner. Commercial salesmen aijd ac¬ 
countants can make out a net list of the series likely to be used, and their 
h»bor will be greatly facilitated. It is immaterial which per cent of the 
series is used first, 60 and 5 off being the same as 5 and 60 off, etc. 


RAPlp MiJltIplIcatIoW. 



PRACTICAL GEXEKAL METHODS. 

The method known as cross multiplication, by 
which the partial products are added mentally, has been 
known by mathmeticians for ages, and while the rule is used 
practically by nearly every really expert accountant, it has 
received comparatively no attention from teachers, who do 






























J 


22 THE SHORT-RULE ARITHMETIC. 


not realize the importance of saving time in large commer¬ 
cial establishments. The reason of this neglect on their 
part can also be accounted for by the fact of the rule requir¬ 
ing more mental work than the ordinary method, and being 
more perplexing especially where the multiplier and multi¬ 
plicand each consists of several figures. Many authors of 
arithmetics recommend the method to be taught in the 
schools in cases where the figures of the multiplier do not 
exceed two or three places, but further than that they are 
silent, and present no general remedy except to return to 
the ordinary rule. We will speak of the rule further on, 
together with some valuable improvements in connection 
with it. We will first show an original and easy method of 
multiplying which is quite general in its application and 
saves writing many figures: 


Short Method. 

Multiply 2 9 5 

By 7 3 

88 

2 1 5 ¥5 


Ordinary Method. 

295 
7 3 


885 
20 6 5 


21535 


Explanation.— We draw two lines below the example with space 
enough between them in which to write as many lines of figures less 
one as there are significant figures in the multiplier. In this example, 
as there are two figures in the multiplier, we draw two lines below with 
space enough between them to write one line of figures. We first say 3 
times 5 are 15. Write the 5 below the lower line and carry the 1 to the 
product of 3 times 9, which gives 28. Write 8 between the lines, one 
place to the left (or always under the figure of the multiplicand being 
used), and carry 2 to the product of 3 times 2, which gives 8. Write the 
sum between the lines to the left. Next we simply multiply 295 by the 7 
(tens) as in ordinary multiplication, taking care to add the figures be¬ 
tween the lines in their proper order, as follows: 7 times 5 are 35 + 8 (be¬ 
tween the lines), which gives 43. Write 3 in the answer and carry 4 to 
the product of 7 times 9, which gives 67, to which we add the next figure 
(8) between the lines, making 75. Writing down 5 and carrying 7 to the 
product of 7 times 2 we get 21, which when written in the answer com¬ 
pletes the product. The figures between the lines can be written above 










MULTIPLICATION. 


23 


the example, as in the illustration, without drawing an additional line 
below, but it is preferable, especially in larger numbers, to write the 
extra figures between the lines. 


Short Method. 

Multiply 3 4 7 6 

By 5 1 a 


Ordinary Method. 

3476 
5 12 


695 
3 48 


1779712 


6952 

3476 

17380 


1779712 


Explanation.—As we have three figures in our multiplier we leave a 
space between the lines sufficient to write two lines of figures. We 
first multiply 8476 by 2, which gives 6952. Write the unit figure (2) of 
the product below the lines and the remaining figures one place to the 
left between the lines. We next multiply 8476 by 1, which gives 3476, 
the unit figure of which we do not write, but add it to the first figure (5) 
between the lines, as follows: lX6=6+5 (first figure between the lines) = 
11. Write down the 1 below, and carry 1 to the product of 1X347, which 
gives 848. Write it one place to the left of the other line. We next say 
5 times 6=30+ 8+9 (figures between the lines) =47. Write down 7 and 
carry4. Next,5times7=35 + 4 (tocarry) +4+6 (betweenlines) =49. Write 
down 9 and carry 4. Next, 5 times 4=20+4 (to carry) + 3 (between lines) 
=27. Write down 7 and carry 2 to the product of 5 times 8=17, which 
when written below completes the product. Proceed in like manner 
with any number of figures. 


EXAMPLES FOR PRACTICE. 



174 

1547 

86 

47 

285 

836 

87 

928 


60 

139 

464 

4042 

49590 

1293292 


This method saves writing, on an average, more than half 
the figures of the partial products and requires no extra 
mental effort. As it is general in its application to all num¬ 
bers we see no reason why it should not be acquired by 
every ambitious student. 














24 


THE SHORT-RULE ARITHMETIC. 


IMPROVED CROSS MULTIPLICATION. 

The method of cross multiplication, while requiring con¬ 
siderable mental effort where large numbers are used, is 
quite simple and easy when the multiplier contains only 
two figures. As the calculations of actual business usually 
consist of small numbers it will repay any one to learn to 
multiply two figures by any number of figures in a 
single line. After this has been mastered we arrange a 
method of relieving the mind of adding the partial products 
mentally when larger multipliers are used. 


Short Method. 

Multiply 5 4 

By 4 2 

226$ 


Ordinary Method. 
54 
42 


108 
2 16 


2268 


Explanation.— First, 2 (units) X4 (units) =8. Write it in the answer. 
Next, 2 (units) X5 (tens) =10, to which add the product of 4 (tens) by 4 
(units), which gives 26. Write down 6 and carry 2 to the product of 4 
(tens) by 5 (tens), which gives 22 and completes the product. 


63 

52 


32 76 


EXAMPLES FOR PRACTICE. 

86 94 68 

41 23 53 


3526 2162 3604 


Short Method. 

Multiply 7 3 6 5 

By 4 2 


Ordinary Method. 

7 365 
42 


309330 


14730 

29460 


309330 















MULTIPLICA TION. 


25 


Explanation.— The first and second figures of the answer are found 
as in the preceding examples. We next have 2X3 + 4X6=30, to which we 
add the carrying figure (3), making 33. Write down 3 and carry 3. 
Next, 2 X7 + 4 X 3=26. 26 + 3 (carrying figure) =29. Write down 9 and 
carry 2. The units figure (2) of the multiplier having been multiplied 
by the 7, or last figure of the multiplicand, it is not used further. Next 
multiply 4 by 7, to which we add the carrying figure (2), which gives 30 
and completes the product. 


EXAMPLES FOR PRACTICE. 

325 7324 12345679 

64 93 18 

20800 681132 222222222 

If the student will practice adding pairs of figures, as in 
the grouping method of addition, this, in connection with 
these rules, is the whole secret of rapid multiplication. We 
will now show how to multiply any number by three 
figures, making only one line of figures besides the answer. 


Short Method. 

Multiply 9 0 4 

By 8 3 2 


3 08 


802048 


Ordinary method. 

9 64 
832 


1928 
2892 
7 712 


802048 


Explanation.— We first multiply 964 by 32, as in the preceding exam¬ 
ples, but instead of writing the entire product below the second line we 
only place the tens and units figures there (48), and the remaining figures 
(308) of the product we write one place to the left between the lines. 
We next multiply 964 by the hundreds figure (8) and add the figures be¬ 
tween the lines in their regular order, as follows: 8X4=32, to which we 
add the figure (8) between the lines, which gives 40. Write down 0 and 
carry 4. Next 8X6=48 + 4 (carrying figure) + 0 (between lines) =52. Write 
down 2 and carry 5. 8X9=72 + 5 (carrying figure) + 3 (between lines) =80, 
which completes the product. 











26 


THE SHORT-RULE ARITHMETIC. 


EXAMPLES FOR PRACTICE. 


817 

623 

187 

508991 


9134 

752 


4749 

6868768 


83163 

942 

34928 

78339546 


The product of any number of figures by four figures 
can be obtained by multiplying in pairs or by making two 
lines of partial products, which is much easier, as follows: 


Short Method. 

Multiply 0 7 18 3 

By 9 4 5 3 

3 4 9 3 5 
30873 

035013710 


Ordinary Method. 

0 7 18 3 
9 45 3 


134300 

335915 

308733 

004047 

035013710 


Explanation.— We leave 6pace between the lines for two lines of 
figures. 67183 X 52 (same as in preceding examples) =8493516. Write 16 
in the answer and 84935 one place to the left between the lines. Next, 
67183X4=268732. Add the units figure (2) to the first figure between the 
lines (5) we get 7, which should be written in the answer and the remain¬ 
ing figures, 26873, written between the lines one place to the left. We 
next multiply 67183 by 9, to which we add as we proceed the figures be¬ 
tween the lines, commencing, as previously explained, with those over 
the first vacant space in the answer, as follows: 9 X 3= 27 . 27 + 3 + 3 (be¬ 
tween lines) =33. Write down 3 and carry 3 to the product of 9X8, which 
gives 75. 75+7+9 (between lines) =91. Write down 1 and carry 9 to the 
product of 9X1, which gives 18. 18+8+4 (between lines) =30. Write 
down the 0 and carry 3 to the product of 9X7, which gives 66. 66 + 6 + 3 
(between lines) =75. Write down 6 and carry 7 to the product of 9X6 
which gives 61. 61+2 (between lines) =63, which when written beloyp 
completes the product, 
















DIVISION. 


27 


EXAMPLES FOR PRACTICE. 


8534 

5326 


19438 

8951 


2218 

2561 


9913 
1 7494 


45452084 


173989538 


The beauty and value of this system of rapid multiplica¬ 
tion is apparent to any person who desires to occupy a posi¬ 
tion in business above the ordinary. It is general in its 
application to all numbers, and any child who can master 
the usual method of long division can certainly acquire it. 

Note.—F or special rules in multiplication see page 30. 


AppIrtG METHtop of plVlsloN. 


The ordinary method of long division requires us to write 
down the product of the divisor by each answer figure, which 
is quite lengthy and tedious. The method of subtracting 
each product mentally has never been popular in this 
country, although in France and other countries it is the 
almost universal method. Subtracting the products mentally 
in this manner is quite difficult, as the processes of subtrac¬ 
tion and multiplication are so closely interwoven that con¬ 
fusion is liable to result in carrying. By the adding 
method the subtractions are made by addition, as ex¬ 
plained under the head of subtraction. It will be found 
quite an easy process, and as it requires only about one-half 
the figures of the ordinary method it should be acquired aud 
used by every accountant. 










28 


THE SHORT-RULE ARITHMETIC. 


As this method differs from the ordinary method only in 
the process of subtraction the entire principle is embodied 
in the following example: 

Divide 789 by 91. 


Ordinary Method. 

9 1 ) 789(8 
738 


Adding Method. 

9 1 ) 789(8 


0 1 Rem. 


6 1 Rem. 


Explanation.— We observe that 91 is contained in 789 eight times, so 
we write 8 in the answer, multiply the divisor by it, and place below 
the figures of the dividend being used the figures which when added 
to the product will equal the dividend. First, 8X1 (divisor figure) =8. 
1 added to 8=9 (first dividend figure). Write down the 1. Next, 8X9 
(divisor figure) =72; 6 added to 72=78. Write down the 6, which com¬ 
pletes the remainder. 

While the result is the same there is quite a difference in 
the mental work of finding a figure which will make a number 
equal another number and subtracting one number from 
another. When this principle is applied to long division we 
think of multiplying and adding instead of multiplying and 
subtracting, which are two separate processes. 


Divide 8742 by 61. 


Adding Method. 


Ordinary Method. 


61 ) 8743(143 


6 1 ) 8743(143 


364 

363 


6 1 


1 9 Rem. 


364 

344 


303 

183 


1 9 Rem . 


Explanation.— We first observe that 61 is contained once in 87. 
Write 1 in the answer. 1X1=1; O added to 1=7 (dividend figure). 
Write down the 6. 1X6 (divisor figure) =6. £ added to 6 (dividend figure) 










DIVISION. 


29 


=8). Write down 2, which leaves a partial remainder of 26. Bring 
down 4 (dividend figure), and we have 264. 61 is contained four times 
in 264, write 4 in the answer and proceed as before. 4X1 (divisor figure) 
=4; 0 added to 4=4 (dividend figure), so we write down 0. 4X6 (divisor 
figure) =24. Si added to 24=26, so we write down 2. We now have a 
partial remainder of 20, to which we annex the next figure of the divi¬ 
dend (2), and we have 202. 61 is contained in 202 three times. Write 3 
in the answer. 3X1 (divisor figure) =3. It will now be observed that 
the first right-hand figure of the partial remainder is 2, which is smaller 
than our three obtained in this manner, so, according to the method of 
subtraction by adding (see page 18), we must add the least figure to 3 
which will make the unit figure of their sum equal 2. 9 is the figure 
which when added to 3=12. We write down 9 and carry 1. 3X6 (divisor 
figure) =18, to which we add the 1 (carrying figure), which gives 19. 1 
added to 19=20 (figures above in partial remainder). Write down 1 be¬ 
neath and we have the complete remainder (19). 

9 

As the method of producing each partial remainder is the 
same, we will avoid tiresome repetition by taking one more 
example with only one figure in the quotient, which should 
make the method plain to any one. 


Divide 78287 by 8465. 


Adding Method. 

8465)78387(9 
3 10 3 Rem. 


Ordinary Method. 
8465)78387(9 
76 185 


3 10 3 Rem. 


Explanation.— We assume 8465 to be contained in 78287 nine times. 
9x5=45. Si added to 45=47. Write down 2 and carry 4. 9X6 (divisor 
figure)=54 +4 (carrying figure)=58. © added to 58=58. Write down 0 
and carry 5. 9X4 (divisor figure) =36+ 5 (carrying figure)=41. 1 added 
'to 41=42. Write down 1 and carry 4. 9X8 (divisor figure)=72 +4 (carry¬ 
ing figure)=76. Si added to 76=78 (above), so write down 2, making a 
remainder of 2102. Perform the following example, getting the partial 
remainders in the same manner 





THE SHORT-RULE ARITHMETIC. 


SO 


Adding Hletliod. 

7 13 ) 87574(132 
1637 
2 134 

7 10 Rem. 


EXAMPI.ES 

53)87642(1653 
346 
284 
192 
3 3 Rem. 


Ordinary Method. 

712)87574(122 
7 12 


1637 

1424 


2 134 
1424 


7 10 


PKACTICE. 

956)76265(79 
9345 
7 4 1 Rem. 


Special Short; JfUles iij Multipli¬ 
cation ai)d pivisioi). 


In addition to the rules previously shown, which are 
general and universal in their application, there is quite a 
number of special short rules in multiplication and division 
which depend upon the relations of certain numbers to the 
base of our system of notation and its multiples. Promi¬ 
nent among these are the aliquot parts of 10,100,1000, etc., 
which, owing to custom and convenience, enter largely into 
transactions of actual business. 

The following table shows most of the aliquot parts used 
in business, and should be mastered as thoroughly as the 









SPECIAL SHORT RULES. 


31 


multiplication table. Simply memorize the aliquot parts of 
100, and the aliquot parts of 10 and 1000 are either one-tenth 
or ten times the corresponding aliquot part of 100: 


TABLE OF ALIQUOT PARTS. 


60 = one-half of 100. 

331 = one-third of 100. 

66 j = two-thirds of 100. 

25 = one-fourth of 100. 

75 = three-fourths of 100. 

16i = one-sixth of 100. 

83£ = five-sixths of 100. 

12i = one-eighth of 100. 

37i = three-eighths of 100. 

62i = five-eighths of 100. 

87 i = seven-eighths of 100. 

8£ = one-twelfth of 100. 

41£ = five-twelfths of 100. 

58i = seven-twelfths of 100. 

91& = eleven-twelfths of 100. 

61 = one-sixteenth of 100. 

181 = three-sixteenths of 100. 

311 = five-sixteenths of 100. 

43f = seven-sixteenths of 100. 

561 = nine-sixteenths of 100. 

681 = eleven-sixteenths of 100. 


811 = thirteen-sixteenths of 100. 

931 = fifteen-sixteenths of 100. 

Note.— Decimal Point— Before proceeding further the student should 
thoroughly understand the leason for removing the decimal point, as it 
is of the utmost importance to a thorough knowledge of aliquot parts 
as applied to business. The point is always at the left of a decimal and 
on the right of a whole number. If the whole number has no decimal 
attached-being frequently omitted for convenience—it must be borne 
in mind whether the point is written or not that it is presumed to be 
on the right of a whole number, and that only one point appears in one 
number either whole, decimal, or mixed. The point must never be 
omitted in a decimal. Removing the decimal point to the right mul¬ 
tiplies a number in a ten-fold ratio, and removing it to the left 
divides a number in a ten-fold ratio. Therefore if we wish to divide 
by 10 we simply remove the point one place to the left, by 100 two places, 




82 


THE SHORT-RULE ARITHMETIC. 


by 1000 three places. For instance, we wish to divide 647 by 10. Here it 
is observed the point is not expressed, but it is presumed to be on the 
right of 647, so we remove it one place to the left, which gives 64.7. If 
we divide the same number (647) by 100 we get 6.44, and by 1000 we get 
.647. Now again, suppose we wish to divide the same number by 10000, 
we remove the point four places to the left. As there are only three 
figures in the number (647) we must prefix a cipher to complete the 
number of places and place the point to the left, which gives .0647. Pro¬ 
ceed in like manner with any number of places. The rule when divid¬ 
ing or multiplying by 10, 100, or 1000 is to always remove the decimal 
point as many places as there are ciphers in the divisor or multiplier, as 
the case may be. Now again, suppose we wish to multiply 647 by 10. 
As 647 is a whole number we simply annex a cipher, which gives 6470; 
647X100=64700, etc. Next, suppose we wish to multiply 647.53 by 10. 
We simply remove the point one place to the right , which gives 6475.3, 
and so on to any number of places. After thoroughly understanding 
how to multiply and divide by 10, 100, 1000, etc., in this manner, it be¬ 
comes an easy task to multiply or divide by any aliquot part. 

Divide 888 by 25. 

Operation. 

8.88 

4 

3 5.52 Answer. 

Explanation.—W e first divide 888 by 100, which gives 8.88. As there 
are four 25s in 100 there are four times as many 25s in 888 as there aro 
hundreds, which gives 35.52, answer. 

Divide 888 by 33$. 

Operation. 

8.88 

3 

2 6.64 Answer. 

Divide 321.56 by 16$. 

Operation. 

3.2156 

6 


19.2936 Answer. 







SPECIAL SHORT RULES. 


33 


Proceed in like manner with any aliquot part. Besides 
those mentioned in the table there are several special cases 
where numbers are not aliquot parts but being so near to 
them that they are worked on that basis, which renders the 
answer sufficiently exact for all practical purposes. A perch 
of stone contains 24f cubic feet, and in some localities 16i 
cubic feet. Either of these numbers are near convenient 
aliquot parts; 24£ being nearly 25, or one-fourtli of 100, and 
16 h nearly 16£, or one-sixth of 100. 

Suppose we wish to find the number of large and small 
perches of stone in a wall which contains 1150 cubic feet. 


Operation, 


Operation, 


11.50 

4 


11.50 

6 


46.00 

.46 


69.00 

.69 


4 6.4 6-{-Large perches. 69.6 9 -\-Small perches. 

Explanation.— We first divide 1150 by 100 and multiply by 4, which 
gives (46) the number of times 25 is contained in the number. Our 
divisor is 24J. By adding rfo, of the quotient (46) to itself we get 46.46 
perches, which is true to the second decimal figure. With small perches 
164 is so near 161 that we divide by 164, as shown above, and add T hs of 
the quotient, as with large perches. If greater accuracy is required the 
answer can be easily found to any required place by continuously re¬ 
moving the quotient two places to the right and adding the results. 

The number 144 occurs as a divisor quite frequently, 
especially in square and cubic measure and in buying and 
selling by the gross. Thousands of different kinds of arti¬ 
cles of hardware, notions, etc., are sold at wholesale by the 
gross. As 144 is just a little more than one-seventh of 1000, 
multiplying the cost per gross in cents by .007 affords a 
valuable short rule for finding the price of a single article 
in cents. A little practice will enable the student to get the 
answer mentally in an instant. 


3 








34 


THE SHORT-RULE ARITHMETIC. 


Purchased a lot of tooth brushes at $19.00 per gross. 
Find the cost of a single brush. 


Short Method. 


1 



Gross price in 
cents-t-lOQQ. 


13 + Cents apiece. 


Ordinary Method. 

144)19.0 0(.18 + Cents apiece. 
144 

460 

432 

28 


Explanation.— As 144 is a little more than one-seventh of 1000 the 
answer will always be slightly too large, bat on articles that cost as high 
as $144.00 per gross the error will not amount to a cent. In business 
the object is always to get the answer to the nearest cent, and therefore 
a rule of this kind, while not mathematically correct, will answer every 
requirement. 


When dividing by such numbers as 60, 700, 2000, etc., 
never commit the folly of carrying the ciphers through the 
calculations by long division. A few examples will make 
this clear. 

In 43731 pounds of wheat how many bushels? 

Operation. Operation. 

60)4373.1 or 60)4373.1 

7 2 8 fi Bus. 7 2 8.85 Bus. 

As there are 60 pounds in a bushel of wheat our divisor is 60. Remov¬ 
ing the point one place to the left in the dividend we get 4373.1 the 
number of times 10 is contained. As there are six tens in 60, one-sixth of 
the number of tens is the answer. Simply divide by 6 until the decimal 
point is reached, where we have a remainder of 5, to which we annex 
the figure (1) on the right of the decimal, which gives 51, the remainder 
or number of odd pounds. It can be carried through by decimals in 
the same manner, as shown on the right. 

In 53345 pounds of hay how many tons of 2000 pounds 
each? 










SPECIAL SHORT RULES. 35 


Operation. Operation. 

2000)53.345 or 2000)53.345 

2 6 Tons. 2 6.6725 Tons. 

Removing the point three places to the left divides by 1000. Simply 
divide this quotient by 2 and we have the number of times 2000 is con¬ 
tained. Always remove the point to the left as many places as there are 
ciphers on the right of the divisor. 

Divide 98746 by 4300. 

Operation. Operation. 

4300)987.46(22 or 4 3 00)987.46( 22.96 + 

127 127 

41 414 

276 

18 

Explanation.— As there are two ciphers on the right of the divisor 
we remove the point two places in the dividend, and as the divisor con¬ 
tains more than one significant figure (43) we must employ long division 
(see adding method of division on page 27). We divide until we come 
to the point in the dividend and find we have a partial remainder of 41, 
to which we annex the figures on the right of the point in the dividend 
which gives 4146, the full remainder. The answer is obtained in deci¬ 
mal form in example on the right. 

MULTIPLYIXG BY ALIQUOT PARTS. 

(See table page 31.) 

Multiply 743 by 25. 

Operation. 

4)74300 

18 5 7 5 Answer. 

Explanation.—B y annexing two ciphers to 743 we multiply it by 100. 
As we wish to multiply 743 by 25, or one-fourth of 100, we simply take 
one-fourth of 74300=18575, answer. 

EXAMPLES FOR PRACTICE. 

Multiply 9576 by 33£. Multiply 4325 by 16£. 

Operation. Operation. 

8)957600 6)432500 


819 2 0 0 Answer, 


7 2 0 8 31 Answer . 









36 


THE SHORT-RULE ARITHMETIC. 


In addition to the method of aliquot parts there are sev¬ 
eral short special rules in multiplying which deserve atten¬ 
tion. 

To multiply by any number of similar figures, as 11, 111, 
4444, 88888, etc.: 

Multiply 5 3 24 
By _11 

5 8 5 6 4 Answer. 

Explanation.— Write the units figure (4) of the multiplicand in the 
answer. Add the units (4) and tens (2) and write their sum (6) in the 
answer. (If the sum is more than 9 we place down the unit figure of 
the sum and carry the tens.) Next, the sum of tens (2) and hundreds (3) 
is 5. Write it down. Next, the sum of hundreds (3) and thousands (5) 
is 8. Write it down, and bringing down the thousands we have 58564, 
answer. 

Multiply 5 3 2 4 

By 111 

5 9 0 9 6 4 Answer. 

Explanation— First bring down the 4. Next, 4+2=6. Write it 
down. It must be observed that as we have three figures in the multi¬ 
plier we must now add together three figures of the multiplicand until 
we get to the extreme left, as follows: 4+2+3=9. Write it down. Now 
drop the first figure (4), and add the next three figures. 2 + 3 + 5=10. 
Write down 0 and carry 1. Next we say 3 + 5=8 +1 (carrying figure) =9. 
Write it down. Next, bring down the last figure (5) and the product is 
complete. 

Multiply 5 3 2 4 

By 1111 

5 914 9 6 4 Answer. 

Explanation.— First bring down 4. Next, 4+2=6. Next, 4+2 + 3=9. 
We now have four figures in the multiplier, and consequently must now 
add four figures of the multiplicand, as follows: 4 + 2 + 3 + 5=14. Write 
down 4 and carry 1. Next, drop first figure (4) and we have 2 + 3 + 5=10, 
together with carrying figure, we have 11. Write down 1 and carry 1. 
Next, 3+ 5=8 + 1 (to carry)=9. Next, bring down last figure and the 
product is complete. 

Multiply 5 3 2 4 
By 7 7 7 7 

5914964 

7 


41404748 Answer. 








SPECIAL SHORT RULES. 


37 


Explanation.— We first multiply as in the preceding example, which 
gives the product of 5324 by 1111. As 7777 is seven times 1111, we simply 
multiply by one 7. If the multiplier was composed of 8s we would 
multiply by one 8, etc. 

The product of any number by any number of nines , like 
9, 99. 999, etc., can be obtained in two ways by subtraction , 
as follows: 

Multiply 842675 by 999999. 

First Short Method. Second Short Method. 

842675000000 or 999999 

842675 842675 

842674157325 842674157325 


Explanation.— By the first process we annex as many ciphers to the 
multiplicand as there are nines in the multiplier and subtract the multi¬ 
plicand from the number thus obtained. 

The second process is preferable when the factor which contains the 
nines is equal to or larger than the other factor. In 6uch cases subtract 
the other factor, less one , from the number of nines , and to the re¬ 
mainder thus obtained prefix the same factor, less one. 


To multiply two figures by two figures whose units sum 
is ten and tens are alike. 

Multiply 96 by 94 and 65 by 65. 


Operation. 

96 

94 


9024 


Operation. 

65 

65 


4225 


Explanation.— Take the first example. 4(units)X6 (units) =24. Write 
down 24. Add 1 to either tens figure, which gives 10, and multiply it by 
the other tens figure, which gives 90, and when prefixed to 24 = 9024, 
answer. When the product of the units by the units does not exceed 9, 
fill the tens place of the answer with a cipher, as for instance, 89X81= 
7209, etc. 

This principle can be extended to all numbers of two 
figures whose units add to 10 and whose tens figures are 
unlike , as follows; 







38 


THE SHORT-RULE ARITHMETIC. 


96 

84 


85 

75 


78 

52 


93 
4 7 


8064 


63 75 


4056 4371 


Explanation.— Take the first example we say 4 (units) X6 (units)=24. 
Write down 4 and carry 2. Next substract the smaller tens (8) fron the 
larger tens (9) which gives 1. Multiply the remainder (1) by the units 
figure (4) of the smaller factor (84) which gives 4, to which we add the 
carrying figure (2), which gives 6. Write it down. Next add 1 to the 
upper ten and multiply by the lower, which gives 80 and completes the 
product. This also applies to mixed num bers the sum of whose fractions 
equal a unit. 

There are hundreds of other special short rules in multi¬ 
plying and squaring numbers, but we will not include them 
here. Rules similar to the one precediDg are of but little 
practical value, except when the student is on the lookout 
for every little mental help, and in teaching, if presented 
properly instead of confusing, they will not fail to interest 
the pupil. 

Before learning the special rules we advise careful and dil¬ 
igent study of the general short methods which are the 
prime features of this work. 


Contracted Method of Mu)tip)ica« 
tioij of peciijials.* 


It seldom occurs that strict accuracy in the answer of a 
business example is required further than cents or hun¬ 
dredths place; therefore the old method which compels us 
to produce unnecssary figures to obtain the necessary ones, 
is a waste of time and mental labor. Examples containing 


* Note.—T he method we present here is far superior to the old form 
of this rule which required us to reverse one of the factors, which made 
it very objectionable for teaching or for business. 










MULTIPLICATION OF DECIMALS . 39 


whole numbers only are few in proportion to those involv¬ 
ing fractions, and the fractions of business are not of large 
denomination or of great variety. Common fractions of 
bushels, pounds, yards, feet, etc., of various commodities, 
and their respective prices can be easily changed into deci¬ 
mal form or memorized (see table of aliquot parts, page 31), 
after which by this process of multiplication only such por¬ 
tions of either factor necessary to produce the answer suffi¬ 
ciently exact need be used. The operations of interest, dis¬ 
counts, averaging accounts, measurements, etc., are wonder¬ 
fully facilitated. 

Find the product of 432.56 by .7 to the nearest hundredth. 

Operation. 

432.56 

.7 

3 0 2.7 9 -{-Answer, 

Explanation.— By this method we always place the point of the mul¬ 
tiplier directly beneath the point of the multiplicand. Be sure to re¬ 
member that tenths place in both factors is our guide or dividing line. 
The tenths figure of the multiplicand and all the figures to its left are 
multiplied by the tenths figure of the multiplier as in ordinary multipli¬ 
cation. Thus in this example we have 4325X7 to the product of which 
however, we must add the carrying figure (4) which is procured men¬ 
tally by multiplying 7 by 6 (the hundredths or rejected figure of the 
multiplicand). It is immaterial where the answer is written beneath as 
we always have two figures to point off. It is best, however, to preserve 
the form indicated in these examples as the points of the partial pro¬ 
ducts will fall under those of the multiplier and multiplicand i 

Find the product of 432.56 by .07 to the nearest hundredth. 

Operation. 

432.56 

.07 


3 0.2 8— Answer, 

Explanation. —We now observe that the tenths place of the multi¬ 
plier is 0, but it serves as our guide in every instance. We now simply 





40 


THE SHORT-RULE ARITHMETIC. 


multiply 432 by 7, but first we get our carrying figure from the first re¬ 
jected figure (5). We say 7X5=35. Before proceeding further let us 
dwell upon the importance of getting these carrying figures equalized, 
because the correctness of our product in a long example may be uncer¬ 
tain unless we use proper judgment as we proceed. It was seen in the 
preceding example that our carrying figure was 4 or the tens figure of 
the product of 7X6. We took the 4 because 42 is nearer 40 than 50. 
Now in this case 35 is midway between 30 and 40, and we must always 
give the upper number preference, especially if there are figures on the 
right of the rejected figure. Therefore we will now call the carrying 
figure 4 instead of 3. So we simply say 432 multiplied by 7 to which we 
add the carrying figure. The answer 30.28 is slightly too large, but it is 
to the nearest hundredth. 

Find the product of 432.56 by .007 to the nearest hundredth. 

Operation. 

432.56 

.007 

3.0 3— Answer. 

Explanation.— The first 0 rejects the 5 above, and the second 0 rejects 
the 2 above, consequently we have 43X7 to which we add 2 (to carry). 
The ingenious student can look back over the rejected figures 256 and 
perceive that the carrying figure is really nearer 2 than 1. A little good 
mental drill on these carrying figures is absolutely necessary. 

Find the product of 432.56 by .0007 to the nearest hun¬ 
dredth. 

Operation. 

432.56 

.0007 

. 3 0-f Answer. 

Explanation.— First 0 rejects 5, second 0 rejects 2, third 0 rejects 3, 
consequently we say 7X4=28, to which we add carrying figure (2), which 
gives .30, answer. 

Find the product of 432.56 by .00007 to the nearest hnn- 
dredth. 

Operation. 

432.56 

.00007 


0 3—Answert 







MULTIPLICATION OF DECIMALS. 


41 


Explanation.— First 0 rejects 5, second 0 rejects 2, third 0 rejects 3, 
fourth 0 rejects 4, but as our carrying figure is always hundredths we 
must not think our answer is 0. 7X4=28, which is nearer 30 than 20, so 
we carry 3, making 3 hundredths, answer. 

If the student will closely follow these instructions, with 
some practice it will be found an easy road. 

We will now embody all these separate explanations in 
one example which will show the full working of the rule. 


Multiply 432.56 by .77777 to the nearest hundredth. 


Short Method. 

4 3 2.5 6 
.7 7 7 77 


Ordinary Method. 

4 3 2.5 6 
.7 7 77 7 


3 0 2 7 9 
3028 
303 
30 
3 


302792 

302792 

302792 

302792 

302792 


3 3 6.4 3 


3 3 6.4 3 2 1 9 1 3 


Multiply 432.56 by 7. 

Operation. 


432.56 

7. 


3027.92 

Explanation.— This example is given to show that when we have 
whole numbers in the multiplier, the tens figure goes by the hundredths 
of the multiplicand and all on its left, etc, It must be remembered that 
this rule does not save figures when only whole numbers occur, but it 
avoids useless figuring when decimals are involved. 

We will now present a few appropriate examples to illus¬ 
trate the practical value of the rule. 

Find the cost of 17f yards of cloth @ 13^ per yard, 
and 91f pounds of butter @ 7£^ per pound. 










42 THE SHORT-RTJLE ARITHMETIC . 


Operation. 

Operation. 

1 7.6 2 5 

91.7 5 

.13 5 

.0 7 75 

176 

642 

53 

64 

9 

5 

$ 2.3 8 —Answer. 

$7.11 -\-Answer. 

The interest of SI @ 8% for 47 days is .00783+ * Find the 
interest on S93.25 and $941.75 at the same rate and time. 

Operation. 

Operation. 

$93.25 

.00783 

$ 9 41.7 5 

.00783 

65 

7 

659 

75 

3 


$ . 7 2-J-^.ttSwer. - 

$ 7.3 l-\-Answer. 


The circumference of any circle is equal to the diameter 
multiplied by 3.14159+; find the circumference of a circle 
whose diameter is 5.85 inches. 

Operation. 

3.14159 
5.8 5 


1571 

251 

16 


1 8.3 8—Inches, diameter. 


* For short, methods of getting the interest of $1 at any rate per cent 
for any length of time, see “interest.” 















FRACTIONS. 


43 


FliACTfortS. 


ADDITION OF FRACTIONS. 

Find the sum of f and ^. 

Operation. 

3. jjj>. 

f -f ir — fl Answer. 

Exlanation.— We multiply the denominator (7) of the first fraction 
by the numerator (5) of the second fraction, which gives 35. Next we 
multiply the numerator (3) of the first fraction by the denominator ('ll) 
of the second fraction, which gives 33. We now unite these products 
(35 + 33=68), which gives the numerator of the answer. The denominator 
of the answer is the product of the denominators (7X11=77). 

Find the sum of -f, § and |. 

Operation. 

4 « 63 

f + i + I = W Answer. 

Explanation.— We take the numerator (4) of the first fraction and 
multiply it by the product of the denominators of the other fractions 
(3X4=12). 12X4=48. Write it down. We next multiply the second 

numerator (2) by the product of the right and left-hand denominators 
(7X4=28) 28X2=56. Write it down. Next, the third numerator we 
multiply by the product of the first two denominators (3X7=21), 21X3= 
63, which completes the partial sums. We add them and get 167, which 
written over the product of the denominators (7X3X4=84) completes 
the sum. 


SUBTRACTION OF FRACTIONS. 


From i take f. 


¥ 


Operation. 



& Answer. 





44 


THE SHORT-RULE ARITHMETIC. 


Explanation.— First we multiply the denominator (8) of the subtra¬ 
hend by the numerator (2) of the minuend, which gives 16. Next, we 
multiply the numerator (7) of the subtrahend by the denominator (3) of 
the minuend, which gives 21. We then take the smaller product (16) 
from the greater (21), which leaves 5, the numerator of the answer. The 
product of the denominators (24) is the denominator of the answer. 

From 17f take 12A. 

Operation. 

17 f + A 
12 A 

4 ||- Answer. 

Explanation.— We observe that the fraction of the minuend is larger 
than the fraction of the subtrahend. In such cases add one to the 
minuend and deduct it from the subtrahend. 12 + 1=13. 17—13=4, the 
whole number of the answer. Deduct the fraction of the minuend from 
one and add the difference to the upper fraction, thus: i}—TT=tV i + 
t 2 i=sb> the fraction of the answer. 


MULTIPLICATION OF MIXED NUMBERS 

When the answer is desired in correct fractional form we 
may employ the following general method without reducing 
each factor to an improper fraction: 

Multiply 94} by 63&. 

Operation. 


6 5 8 

9 4 

1 

12 6 

6 3 

£ 

5 9 2 2 

14 


8 2 

6 


4 2 

0 


6 0 4 6 H Answer. 

Explanation.—F irst find the product of the whole numbers, which 
can always be' done in a single line (see page 24). We write the product 








DIVISION OF MIXED NUMBERS. 


45 


(5922) beneath, and to the right of this line write the product of the 
numerators of the fractions (2X7=14). Next multiply the numerator 

(7) of the lower fraction by the upper whole number (94), which gives 
658, which write on the left of the upper number. Now divide the 
product thus obtained by the denominator (8) of the lower fraction, 
which gives 82 and a remainder of 2. Write the 82 in the whole number 
column and the remainder (2) we multiply by the upper denominator 

(8) , giving a product of 6, which we write under 14 in the fraction 
column. Next, multiply the lower whole number (63) by the numer¬ 
ator (2) of the upper fraction, which gives 126. Write it on the left. 
Now divide 126 by the denominator (3) of the upper fraction, which 
gives 42 with no remainder. Write 0 in the fraction column. Now sim¬ 
ply add the partial products and our product is complete. If the partial 
products of the fractions amount to more than 1 carry the excess to the 
whole numbers. 


DIVISION OF MIXED NUMBERS. 

Divide 49* by 9. 


Operation. 

9 )49 A 

5 a Answer. 

Explanation.— In the usual method both dividend and divisor are 
reduced to the same denomination. In cases where the divisor is a 
whole number this is unnecessary. We proceed as follows: 9 is con¬ 
tained 5 times in 49 with a remainder of 4. Writedown 5 ,and to produce 
the fraction of our quotient we multiply the remainder (4) by the denom¬ 
inator (11), which gives 44; to this we add the numerator (3) and we have 
47, the numerator of the quotient, The product of the divisor by the 
denominator is the denominator (99) of the answer. 


EXAMPLES FOR PRACTICE. 

Divide 732 * by 9, and 6857 £ by 12. 

9)732* 12)68571 


81 U 


5 7 1 « 







sIMple Interest 



GENERAL, SHORT METHOD. 


There are a great many special short rules for computing 
interest all hinging upon the relations the time, per cent, 
and principal bear to the year or month assumed as a base. 

It is commonly assumed in computing interest that a 
month contains 30 days and a year 360 days. When the 
principal, time, and rate per cent are prime to the number 
of days or months in a year all special short rules are at a 
( 46 ; 









































INTEREST. 


47 


disadvantage, and really most of them, as shown in calcula¬ 
tors and arithmetics are in many cases longer and more 
tedious than the old method of aliquot parts. 

We will discuss the merits of the best of these further on, 
but will now proceed to explain a method which is practical 
and general in its adaptation to all possible cases and saves 
a vast amount of useless figuring. 

If we could instantly find the interest of $1 at any rate 
per cent for any length of time mentally, further than that 
the calculation would be easy. Per cents such as 61, 7,1L 
etc., are prime to the number of days in a year, and the 
mental operation in producing the interest of SI is too great; 
therefore we assume a base rate which is not prime and 
make corrections from it, more or less, as the case may be. 
The corrections can be made from the interest of SI or from 
the principal. 

In assuming 360 days a year the rates of 3 and its multi¬ 
ples, as 6, 9,12, etc., offer a number of easy routes in treat¬ 
ing the time and producing the interest of SI mentally. 

The most convenient base rate to use is 6, as it is the 
legal rate in most States, and besides, the interest of SI for 
any length of time can be produced mentally in an instant, 
In the first place learn the decimal equivalents of all the 
fractions of sixths , as follows: 


\ = .16 + 

I = .33 + 

f = .5 

I = .66 + 

f = .83 + 

Find the interest of SI for 47 days. 

Explanation.— We divide the time in days by 6, and when the time is 
less than 60 days we place down a point and two ciphers (.00). 47-*-6=7 
and 5 remainder. We write the 7 to the right of the .00 and we have .007, 
to which we annex the decimal equivalent of | (83 +) which gives .00783 + 
laswer. The decimal extended would be .007833333+ . 



48 


THE SHORT-RULE ARITHMETIC. 


Find the interest of @ 6 % for 


13 days. 

Answer $ .00216-f- 

29 “ 

tt 

.00483+ 

17 “ 

tt 

.00283+ 

16 “ 

ti 

.00266+ 

18 “ 

a 

.003 

24 “ 

tt 

.004 

48 “ 

tt 

. .008 

53 “ 

a 

.00883+ 

49 “ 

u 

.00816+ 

37 “ 

it 

.00616+ 

59 “ 

tt 

.00983+ 


Note.—W hen the days are less than 6 the interest is simply one of the 
decimal equivalents of sixths with three ciphers. 

Find the interest of $1 @ 6 % for 73 days. 

Explanation.—W hen the days are less than 600 and more than 60 we 
have only one cipher before writing the significant figures of the inter¬ 
est. 73 days are more than 60, therefore we write down a point and a 
cipher (.0). 73-*-6=12 and a remainder of 1. We annex 12 to the cipher, 
gives .012, to which we annex the decimal of h (.16 +), which gives .01216 + 
answer. 

Find the interest @ 6 % for 


60 days. 

Answer $ .01 

63 “ 

u 

.0105 

79 “ 

tt 

.01316+ 

84 “ 

tt 

.014 

93 « 

tt 

.0155 

314 “ 

u 

.05233+ 

423 “ 

a 

.0705 


When the time 'consists of years, months, and days, we 
have the following: 

Find the interest of $1 for 2 years, 8 months, and 12 days. 

Answer , $.162 






INTEREST. 


49 


Explanation.— We multiply the years by 6 and to the product add 
one-half the number of months. Thus, 2 (yrs.)X6=12. 8 (mos.)+2=4. 
We now add 12 and 4, which gives 16. This result is always cents ; there¬ 
fore we place a point, and on the right place the 16. We now take one- 
sixth of the number of days (12), which gives 2, and when annexed to 
.16 we have $.162, answer. 


Find the interest of $1 @ 6 % for 3 years, 9 months, and 19 
days. Answer , $.22816+ 

Explanation.— In this example the months are odd , and a little hint 
here will enable us to avoid using a fraction. We proceed as in the 
preceding example. We say 3 (yrs.)X6=18. Take one-half of 9 (mos.) 
we have 44. We say 18 + 4=22 (cents), which we write down, and the 
fraction of 4 we call 30 days (or 3 tens), and when added to the number 
of days (19) we get 49. 49+6=8 and a remainder of 1. Write down 8 to 
the right of the .22, and we have .228, to which we annex in decimal form 
the remainder of sixths (1), we have .22816 + , answer. 


Find the interest of $1 @ 6 % for 
5 years, 4 months, and 17 days, 


2 “ 9 

1 “ 7 

3 “ 4 

8 

9 


“ 11 “ 

“ 18 “ 

“ 26 “ 

“ 12 “ 

“ 21 “ 


$.32283+ Answer. 
.16683+ “ 

.098 “ 


.14433+ “ 

.042 “ 

.0485 “ 


When the interest of $1 at any other rate per cent is de¬ 
sired we first find the interest at 6 % and then for each whole 
per cent above 6 we add l more and for each whole per cent 
less than 6 we deduct \ of the interest. Thus, if we desire 
the interest at 7 % we would add £, if 8 %, f or 1, if 9 %, f or |, 
etc. If we add or deduct £ for each whole per cent we 
should add or deduct + for each one-half per cent and 
for each one-fourth per cent, etc. 

Find the interest of $1 @ 7 % for 2 years, 4 months, and 18 
days. 

Operation. 

6 ) . 1 4 3 

.0 2 3 8 3 + 



. 16683 + 







50 


THE SHORT-RULE ARITHMETIC. 


Explanation. —We first find the interest at Q%, as in the preceding 
examples, and we get .143. One-sixth of .143=.02383 + which added to 
,143=.16683 + . Now, suppose we wish to find the interest at 71$. We ob¬ 
serve that there are three halves above 6, consequently we should add & 
or i of the interest @ 6$, etc. 

Find the interest of SI for 


42 

days 

@ 8 % 

Answer $ 

.00933+ 

73 

H 

“ 9 % 

u 

.01825 

213 

U 

“ 

u 

.04137+ 

63 

u 

“ 5 % 

(( 

.00875 

21 

u 

“ 4 % 

a 

.00233+ 

83 

(( 

“ 

« 

.01037+ 


We are now prepared to compute the interest on any sum 
of money. 

Find the interest of $241.50 @ 6 % for 47 days. 

Operation. 

$ 2 41.5 0 

.00783 + Int.of$l. 

169 

19 

1 

$ 1.8 9 + Answer . 

Explanation.—O f course any rule for multiplying the principal by 
the interest of $1 will produce the complete interest, but the contracted 
method of multiplying decimals (see pages 38 to 42) will be of great 
assistance in disposing of useless figures, as our answer need be correct 
only to the nearest cent. As the time of most notes, etc., is less than a 
year the interest of $1 is usually a very easy multiplier by this method, 
no difference how difficult it may appear by the old method of multi dy¬ 
ing decimals. 

We can if we choose always use the interest of $1 @ 6 % as 
a multiplier and add to or deduct from the principal such 
portion of it as the per cent is greater or less than 6. There¬ 
fore, if the student will practice on the method of instantly 
getting the interest of $1 @ 6 % mentally, this, together wit> 





INTEREST. 


51 


the short method of multiplying, will undoubtedly, afford 
the shortest route to the required interest. Of course there 
may be cases where some special short rule which applies 
only to a few examples will produce the answer with fewer 
figures, but in the vast majority of business examples this 
will be found the one to rely upon. 

Find the interest of $248.75 for 1 year, 4 months, and 11 
days at 7k 


Operation. 

$2 48.7 5 
62.1 9 

$ 3 1 0.9 4 

.08183 + 

2487 

31 

25 

1 


$ 2 5.4 4 -f- Answer. 

Explanation. —71 % is 11# greater than 6#. For each half above 6# 
we add A, and as there are three halves above we add & or 1 of the 
principal to itself. One-fourth of $248.75 is $62,181, or nearly $62.19. 
Always add to the nearest cent in this manner. $248.75 + $62.19=$310.94. 
In other words, it will require $310.94 @ 6# to earn as much money 
at interest as $248.75 could at 71#. We find the interest of $1 @ 6£ by 
the preceding rule is .08183 +, which we multiply by the principal—using 
the contracted method of decimals, and we have $25.44 +, answer. 

It is very seldom that a more difficult example than this one is en¬ 
countered in actual business. The idea of making the correction from 
the principal is, we believe, entirely original, and is to be preferred to 
making corrections from the interest. 


EXAMPLES FOB PBACTICE. 

Find the interest of $925.75 @6% for 53 days, and $78.42 
@ 7$ for 1 year, 7 months, and 11 days. 






52 


THE SHORT-RULE ARITHMETIC. 


$ 9 2 5.7 5 

.00883 + 


Operation. 


Operation. 

$ 7 8.4 2 
13.0 7 


740 

74 

3 


91.4 9 


.09683 + 


$ 8.1 7 + Answer. 


823 

55 

7 


3 8.8 5 + Answer. 


ACCURATE INTEREST. 


Accurate interest is based on the true year of 365 days. 
Interest is computed by counting the exact number of days; 
each day’s interest being part of the interest for one 
year, whereas by the commercial year, so generally used in 
this country, the interest for each day is its part. It 
will thus be observed that the accurate interest for any time 
less than a year is slightly less than when 360 days are taken 
as a base. Ordinary interest can be changed to accurate 
interest by deducting its + part. 73 being a very clumsy 
divisor, an easier method is to employ the following: 

The ordinary interest of a certain note amounts to $182.50, 
what is the accurate interest? 


Operation. 

$182.50 

2.50 

$180.00 


1 8 S 


5 5 
1 * 


Explanation.— Write down the number of dollars to the nearest unit 
(183). Next write down 3 times the tens, hundreds, etc., to the nearest 
unit, thus: 18X3=54 +1 (to carry) =55, Next, 7 times the hundreds to the 
nearest unit, 7X1=7+ 5 (to carry) =12. Add these products and deduct 
their sum from the ordinary interest. 











INTEREST. 


53 


Special Short; Interest Jellies. 


DECI3IAL METHOD FOR DAYS. 

(Assumed year 860 days.) 


One-tenth part of the 
principal is the in¬ 
terest at 

| One-hundredth part 
of the principal is the 
interest at 

One-thousandth part 
of the principal is the 
interest at 

4 % for 900 days. 

5 % for 720 days. 

6 % for 600 days. 

7 % for 514 days.* 

74 % for 480 days. 

8 % for 450 days. 

9 % for 400 days. 

10 % for 360 days. 

12 % for 300 days. 

4 % for 90 days. 

5 % for 72 days. 

6 % for 60 days. 

7 % for 52 days.* 
74$ for 48 days. 

8 $ for 45 days. 

9 % for 40 days. 

10 $ for 36 days. 

12 % for 30 days. 

4 $ for 9 days. 

5 % for 7.2 days. 

6 % for 6 days. 

7 % for 5.2 days.* 

74 $ for 4.8 days. 

8 % for 4.5 day8. 

9 % for 4 days. 

10 % for 3.6 days. 

12 % for 3 days. 


Exlanation.— This method is founded upon the principle that if $1 
in 1 year earns 6 cents at 6£, in J of 1 year—or60days—$1 will earn 1 cent; 
therefore, T fo part of the principal is the interest for 60 days, ufo, of the 
principal the interest for 6 days, etc. 


Find the interest of $321.00 @ 9 % for 40 days, and on 
$763.25 for 60 days at Q%. 

Operation. Operation. 

$ 3 | 2 1.0 0 $ 7 | 6 3.2 5 

Explanation.— Take the first example. If $1 earns 9 cents in 1 year 
at 9% in J of 1 year, or 40 days, each dollar will earn 1 cent, or $321 will 
earn 321 cents. We draw a line two places to the left, which divides by 
100, and we have the interest. When the number of days in a year is not 
exactly divisible by the per cent—such as 7,11, etc.—the time it takes $1 
to earn 1 cent, mill, etc., is fractional, and the rule in such cases is 
almost useless. The better plan is to use 6% as a base, and make addi¬ 
tions or deductions as in the “General Method.” 

We have simply to remember that removing the point one place gives 
the interest for 600 days, two places for 60 days, and three places for 6 
days. A line can be drawn through the principal to represent the point. 


* Nearly. 













54 


THE SHORT-RULE ARITHMETIC. 


Find the interest of $468 for 60 days at 7 %. 


$4 


Operation. 

68 . 

78 


$ 5 j 4 6 Answer, 

Explanation.— Drawing a line two places to the left gives the inter¬ 
est for 60 days 6 %. 7 % is b more, so we add b of the interest to itself, 
which gives $5.46, answer. 

Find the interest of $675.36 for 19 days @ 6 $. 


Operation. 

7 5.36 = Interest for 
3 


6 days. 


2 0 2 6 0 8 = Interest for 18 days. 
1 12 5 6 = Interest for 1 day. 


$2 138 64 = Interest for 19 days. 


Explanation.— Drawing the line three places to the left, and we have 
the interest at 6 % for 6 days. We multiply by 3 and get $2.02608, interest 
for 18 days. We lack a day’s interest of completing the answer so we 
take h of the interest for 6 days .67536-^6=.11256, which added to the in¬ 
terest for 18 days we have $2.13 +, answer. Of course it is not necessary 
to get the answer nearer than cents , so we should omit useless decimals 
in actual business. The line can be drawn on the time as well as on the 
principal as follows: 


Find the interest of $600 for 213 days at 6$. 


Operation. 

$21 | 30 

Explanation.—T he interest of $600 for 213 days is the same as the in¬ 
terest of $213 for 600 days, therefore we take of the number of days, 
and we have 21.3 or $21.30, answer. 

Find the interest of $40 for 743 days at 9 %. 


Operation. 

7 | 43 

Explanation.—T he interest of $1 at 9 % for 1 year is 9 cents, and for 
b of 1 year—or 40 days—is 1 cent, therefore the interest of $40 for 1 day is 
1 cent, and for 743 days is $7.43. 









INTEREST. 


65 


SAVINGS BANK. COMPOUND INTEREST TABLE. 

Showing the amount of $1, from 1 year to 15 years , with Compound 
Interest added semi-annually , at different rates. 



Three 
Per cent. 

Four 

Per cent. 

Five 

Per cent. 

Six 

Per cent. 

Seven 

Per cent. 

Eight 

Per cent. 

Nine 

Per cent. 

Ten 

Per cent. 

4 

year. 

$1.01 

$1.02 

$1.02 

$1.03 

$1.03 

$1.04 

$1.04 

$1.05 

1 

year. 

1.03 

1.04 

1.05 

1.06 

1.07 

1.08 

1.09 

1.10 

14 

years. 

1.04 

1.06 

1.07 

1.09 

1.10 

1.12 

1.14 

1.15 

2 

years. 

1.06 

1.08 

1.10 

1.12 

1.14 

1.16 

1.19 

1.21 

24 

years. 

1.07 

1.10 

1.13 

1.15 

1.18 

1.21 

1.24 

1.27 

3 

years. 

1.09 

1.12 

1.15 

1.19 

1.22 

1.26 

1.30 

1.34 

34 

years. 

1.10 

1.14 

1.18 

1.22 

1.27 

1.31 

1.36 

1.40 

4 

years. 

1.12 

1.17 

1.21 

1.26 

1.31 

1.86 

1.42 

1.47 

44 

years. 

1.14 

1.19 

1.24 

1.30 

1.36 

1.42 

1.48 

1.55 

5 

years. 

1.16 

1.21 

1.28 

1.34 

1.41 

1.48 

1.55 

1.62 

54 

years. 

1.17 

1.24 

1.31 

1.38 

1.45 

1.53 

1.62 

1.71 

6 

years. 

1.19 

1.26 

1.34 

1.42 

1.51 

1.60 

1.69 

1.79 

64 

years. 

1.21 

1.29 

1.37 

1.46 

1.56 

1.66 

1.77 

1.88 

7 

years. 

1.23 

1.31 

1.41 

1.51 

1.61 

1.73 

1.85 

1.97 

74 

years. 

1.24 

1.34 

1.44 

1.55 

1.67 

1.80 

1.93 

2.07 

8 

years. 

1.26 

1.37 

1.48 

1.60 

1.73 

1.87 

2.02 

2.18 

84 

years. 

1.28 

1.39 

1.52 

1.65 

1.79 

1.94 

2.11 

2.29 

9 

years. 

1.30 

1.42 

1.55 

1.70 

1.85 

2.02 

2.20 

2.40 

94 

years. 

1.32 

1.45 

1.59 

1.75 

1.92 

2.10 

2.30 

2.52 

10 

years. 

1.34 

1.48 

1.63 

1.80 

1.98 

2.19 

2.41 

2.65 

11 

years. 

1.38 

1.54 

1.72 

1.91 

2.13 

2.36 

2.63 

2.92 

12 

years. 

1.42 

1.60 

1.80 

2.03 

2.28 

2.56 

2.87 

3.22 

13 

years. 

1.47 

1.67 

1.90 

2.15 

2.44 

2.77 

3.14 

3.55 

14 

years. 

1.51 

1.73 

1.99 

2.28 

2.62 

2.99 

3.42 

3.62 

15 

years. 

1.56 

1.80 

2.09 

2.42 

2.80 

3.24 

3.74 

4.32 


TIME AT WHICH MONEY DOUBLES AT 
INTEREST. 


Rate per cent. Simple Interest. 

2 . 50 years. 

2 i . 40 years. 

3 . 83 years 4 months. 

34. 28 years 208 days. 

4 . 25 years. 

44. 22 years 81 days. 

5 . 20 years. 

6 . 16 years 8 months. 

7 . 14 years 104 days. 

8 _‘.. 124 years. 

9 . 11 years 40 days. 

10. 10 years. 

ONE DOLLAR LOANED 100 YEARS at Compound Interest 
would amount to the following sum: 


Compound Interest 
85 years 1 day. 
28 years 26 days. 
23 years 164 days. 
20 years 54 days. 
17 years 246 days. 
15 years 273 days. 
15 years 75 days. 
14 years 327 days. 
10 years 89 days. 
9 years 2 days. 
8 years 16 days, 
7 years 100 days. 


1 per cent.$2.75 

3 per cent.19.25 

6 per cent.340.00 

10 per cent.13,809.00 


12 per cent.$84,675.00 

15 per cent.1,174,405.00 

18 per cent.15,145,207.00 

24 per cent.2,551,799,404.00 










































































56 


THE SHORT-RTJLE ARITHMETIC. 


piffereijce li* Tiijie Table, 


(FIRST YEAR.) 


Jan. 

Feb. 

Mar. 

i- 

Apr. 

sS 

• 

June. 

July. 

► 

a 

V 

f 

r 

Oct. 

% 

0 

< 

• 

b 

t 

1 

32 

60 

91 

121 

152 

182 

213 

244 

274 

305 

335 

2 

33 

61 

92 

122 

153 

183 

214 

245 

275 

306 

336 

a 

34 

62 

93 

123 

154 

184 

215 

246 

276 

307 

337 

4 

35 

63 

94 

124 

155 

185 

216 

247 

277 

308 

338 

5 

36 

64 

95 

125 

156 

186 

217 

248 

278 

309 

339 

0 

37 

65 

96 

126 

157 

187 

218 

249 

279 

310 

340 

7 

38 

66 

97 

127 

158 

188 

219 

250 

280 

311 

341 

8 

39 

67 

98 

128 

159 

189 

220 

251 

281 

312 

342 

9 

40 

68 

99 

129 

160 

190 

221 

252 

282 

313 

343 

io 

41 

69 

100 

130 

161 

191 

222 

253 

283 

314 

344 

11 

42 

70 

101 

131 

162 

192 

223 

254 

284 

315 

345 

12 

43 

71 

102 

132 

163 

193 

224 

255 

285 

316 

346 

13 

44 

72 

103 

133 

164 

194 

225 

256 

286 

317 

347 

14 

45 

73 

104 

134 

165 

195 

226 

257 

287 

318 

348 

15 

46 

74 

105 

135 

166 

196 

227 

258 

288 

319 

349 

16 

47 

75 

106 

136 

167 

197 

228 

259 

289 

320 

350 

17 

48 

76 

107 

137 

168 

198 

229 

260 

290 

321 

351 

18 

49 

77 

108 

138 

169 

199 

230 

261 

291 

322 

352 

19 

50 

78 

109 

139 

170 

200 

231 

262 

292 

323 

353 

20 

51 

79 

110 

140 

171 

201 

232 

263 

293 

324 

354 

21 

52 

80 

111 

141 

172 

202 

233 

264 

294 

325 

355 

22 

53 

81 

112 

142 

173 

203 

234 

265 

295 

326 

356 

23 

54 

82 

113 

143 

174 

204 

235 

266 

296 

327 

357 

24 

55 

83 

114 

144 

175 

205 

236 

267 

297 

328 

358 

25 

66 

84 

115 

145 

176 

206 

237 

268 

298 

329 

359 

26 

57 

85 

116 

146 

177 

207 

238 

269 

299 

' 330 

360 

27 

58 

86 

117 

147 

178 

208 

239 

270 

i 300 

» 331 

361 

28 

59 

87 

118 

148 

179 

209 

240 

i 271 

301 

332 

; 362 

29 

— 

88 

119 

' 149 

' 180 

' 210 

i 241 

272 

1 302 

! 333 

; 363 

30 

— 

89 

120 

• 150 

' 181 

211 

242 

1 273 

l 303 

i 334 

: 364 

31 

— 

90. 

— 

151 


212 

1 243 

» 

m 


365 


Explanation.— Find the difference of time between March 12 and 
October 27 in the same year. Look for March and underneath, opposite 
12 in heavy-face figures, we find 71. Deduct thit from the figures corre¬ 
sponding in the same manner to Octobei 27 (300), and we get 229 days, 
answer. If a date runs into anothe year find the later date in the oppo¬ 
site table and proceed as before, These tables are useful in finding the 
































DIFFERENCE IN TIME TABLE. 


57 


piffereijce iij Tiijie Table. 

(SECOND yeab.) 


Jan. 

1 

a* 

• 

8 

» 

Apr. 

- 1 

Jane. 

July. 

> 

a 

V 

% 

ft 

* 

• 

Oct. 

© 

< 

• 

ft 

• 


366 

397 

425 

456 

486 

517 

547 

578 

609 

639 

670 

700 

1 

367 

398 

426 

457 

487 

518 

548 

579 

610 

640 

671 

701 

2 

368 

399 

427 

458 

488 

519 

549 

580 

611 

641 

672 

702 

3 

369 

400 

428 

459 

489 

520 

550 

581 

612 

642 

673 

703 

4 

370 

401 

429 

460 

490 

521 

551 

582 

613 

643 

674 

704 

5 

371 

402 

430 

461 

491 

522 

552 

583 

614 

644 

675 

705 

0 

372 

403 

431 

462 

492 

523 

553 

584 

615 

645 

676 

706 

7 

373 

404 

432 

463 

493 

524 

554 

585 

616 

646 

677 

707 

8 

374 

405 

433 

464 

494 

525 

555 

586 

617 

647 

678 

708 

9 

375 

406 

434 

465 

495 

526 

556 

587 

618 

648 

679 

709 

io 

376 

407 

435 

466 

496 

527 

557 

588 

619 

649 

680 

710 

11 

377 

408 

436 

467 

497 

528 

558 

589 

620 

650 

681 

711 

12 

378 

409 

437 

468 

498 

529 

559 

590 

621 

651 

682 

712 

13 

379 

410 

438 

469 

499 

530 

560 

591 

622 

652 

683 

713 

14 

380 

411 

439 

470 

500 

531 

561 

592 

623 

653 

684 

714 

15 

381 

412 

440 

471 

501 

532 

562 

593 

624 

654 

685 

715 

16 

382 

413 

441 

472 

502 

533 

563 

594 

625 

655 

686 

716 

17 

383 

414 

442 

473 

503 

534 

564 

595 

626 

656 

687 

717 

18 

384 

415 

443 

474 

504 

535 

565 

596 

627 

657 

688 

718 

19 

385 

416 

444 

475 

505 

536 

566 

597 

628 

658 

689 

719 

20 

386 

417 

445 

476 

506 

537 

567 

598 

629 

659 

690 

720 

21 

387 

418 

446 

477 

507 

538 

568 

599 

630 

660 

691 

721 

22 

388 

419 

447 

478 

508 

539 

569 

600 

631 

661 

692 

722 

23 

389 

420 

448 

479 

509 

540 

570 

601 

632 

662 

693 

723 

24 

390 

421 

449 

480 

510 

541 

571 

602 

633 

663 

694 

724 

25 

391 

422 

450 

481 

511 

542 

572 

603 

634 

664 

695 

725 

20 

392 

423 

451 

482 

512 

543 

573 

604 

635 

665 

696 

726 

27 

393 

424 

452 

483 

513 

544 

574 

605 

636 

666 

697 

727 

28 

394 

— 

453 

484 

514 

545 

575 

606 

637 

667 

698 

728 

29 

395 

— 

454 

485 

515 

546 

576 

607 

638 

668 

699 

729 

30 

396 

— 

455 

— 

516 

— 

577. 

608 

— 

669 

— 

730 

31 


dates of the maturity of notes, etc. For instance, what is the date of 
maturity of a note to be paid 63 days from July 4? We look to July 4 and 
find 185 to which we add 63 and get 248. 248 is in the September column and 
opposite 5, which is the date required. When February 29 occurs between 
two dates count an additional day. 

































58 


THE SHORT-RULE ARITHMETIC. 


SlfORT pilsiJNlESS RllLES. 


MARKING GOODS BOUGHT BY THE 
DOZEN. 


Articles of nearly every description are sold at wholesale 
by the dozen or gross. As purchases are usually made in a 
hurry it becomes a matter of prime importance to the mer¬ 
chant or buyer to be able to instantly find the retail selling 
price of a single article from the dozon or gross price. 


To make 20 divide the cost per dozen by 10 


It 

33* 

tt 

tt 

tt 

9 


tt 

50 

tt 

tt 

tt 

8 


it 

100 

tt 

tt 

tt 

6 


tt 

40 

tt 

tt 

tt 

10, and add 

* itself. 

tt 

35 

tt 

tt 

tt 

10 

i “ 

it 

37* 

tt 

tt 

t* 

10 

* “ 

tt 

30 

tt 

tt 

tt 

10 

* “ 

tt 

25 

tt 

tt 

tt 

10 

& “ 

tt 

12* 

tt 

tt 

tt 

10, and substract & “ 

tt 

16§ 

tt 

tt 

tt 

10 

A “ 

tt 

18| 

tt 

tt 

tt 

10 

& “ 

Bought 

a dozen hats at $37.50 

per dozen. 

What is the 


retail price of each hat at 20# profit? 


Explanation. —The above table should be memorized. To make 20 56 
profit on a single article we simply divide the cost per dozen by 10, 
which is done by removing the point one place to the left, and in this 
case we have $3.75, answer. 


MARKING GOODS BOUGHT BY THE 
GROSS. 

Small articles, such as brushes, combs, hinges, locks, 
notions; etc., are usually sold at wholesale by the gross. 









SHORT BUSINESS RULES. 


59 


See pages 33 and 34 for short method of dividing by 144, 
which will enable any one to get the cost of a single article} 
mentally from the gross price. The per cent of profit can 
then be added. 


MARKING GOODS BOUGHT BY THE 
NEST. 

Tubs, pails, baskets, etc., are sold at wholesale by the 
nest. If we wish to find the proportionate cost of each, pro¬ 
ceed as follows: 

Bought 1 nest containing 5 tubs costing $6. The smallest 
tub contains 1 gallon, the second 2 gallons, the largest 5 
gallons, etc. 

Cost of 1 gallon $ .40. 

Cost of 2 gallons .80. 

Cost of 3 gallons 1.20. 

Cost of 4 gallons 1.60. 

Cost of 5 gallons 2.00. 

Explanation.— We take the sum of the gallons 1, 2, 3, 4, 5 = 15, and 
divide the price by it. $6.00 divided by 15 = 40 cents, the price of 1 gal¬ 
lon. Then simply multiply by the number of gallons in each tub. Al¬ 
ways divide the price of the nest by the number of gallons in all the 
tubs and multiply the quotient by the gallons in each tub. 


MERCHANTS’ RUUE. 

Retailers frequently sell coffee, sugar, etc., at a given 
number of pounds per dollar, and the following will show 
a simple method of determining the proportionate quantity 
to sell for fractions of a dollar. 

A merchant sells sugar @ 91 pounds per dollar. How 
many pounds will 60 cents purchase? 

Explanation.— Simply multiply the number of pounds sold for$l 
by the cents 1 worth desired. Thus, 9.75X.60=5.85 pounds, answer. 
Hundredths and tenths of a pound can be reduced to ounces by multi¬ 
plying by 16. 

For short methods of Trade Discounts, etc., see 
pages 19 to 21, 






60 


THE SHORT-RULE ARITHMETIC. 


Chronology or Goi)ipilling Tiijte- 


The Julian Calendar, or Old Style, established by Julius 
Caesar 46 B. C., was in use in all Christian countries until 
the year 1582, when Pope Gregory XIII decreed that in all 
Catholic countries the Julian Calendar should be abolished 
and a correction of 10 days be made, making the 5th of the 
month of October in that year the 15th; and also that instead 
of reckoning every four years a leap year the centurial years, 
as 1600,1700,1800 should not be leap years unless divisible by 
400. It will thus be seen that 1600 was a leap year according 
to the New Style, but 1700,1800, and 1900 are not. The next 
centurial leap year will be 2000. The New Style was not 
adopted in England and America until 1752. The Old 
Style is still used in Russia. 

According to the Old Style every fourth year is counted 
a leap year and an extra day is added to February. As the 
true solar year does not contain exactly 365i days this error 
was the cause of the adoption of the New Style, which is some¬ 
times called the Gregorian Calendar, but which was really 
devised by Aloysius Lilius, or Luigi Lilio Ghiraldi, a learned 
astronomer and physician of Naples, who died before its 
adoption. It devolved upon Clavius to make all the calcula¬ 
tions necessary for its verification, and by whom it was 
completely developed and explained in a large folio treatise 
of 800 pages, published in 1608. 

It is our mission to strip the long methods, tables, etc., of 
the folio treatise of their dominical letters, technical terms, 
etc., and to present the simplest rules for finding the day of 
the week of any date, Easter Sunday, and the principal 
church feasts depending on Easter, Age of Moon, etc. 

OLD STYLE. 

To find the day of the week corresponding to any date, 
according to the Julian Calendar, or Old Style; 




CHRONOLOGY. 


a 


RULE—To the given year add its one-fourtlipart ( dis¬ 
carding fractions , if any,)—in leap years add its one* 
fourth part less one—the number of days that have elapsed 
since the beginning of the given year , and 5. The ex¬ 
cess of sevens signifies the day of the weeJc—1 indicating 
Sunday , 2 Monday , etc., while a remainder of 0 indicates 
Saturday. 

The execution of Charles I occurred January 30, 1649; 
required the day of the week. * 

Operation. 

16 4 9 = Year. 

412 = Leap years. 

3 0 = Days since beginning of year. 

5 

7)2096 

2 9 9 — 3 = Tuesday. 

Columbus discovered America Oct. 12,1492; required the 
day of the week. 

Operation. 

14 9 2 = Years. 

3 7 2 = Leap years, less one. 

2 8 6 = Days since beginning of year. 

5 

7)2155 

3 0 7 — 6 = Friday. 

Explanation.— When the year is exactly divisible by 4 it is always a 
leap year. Old Style. 1492 being a leap year year we must be careful and 
count the extra day for February; also to deduct 1 from the leap years. 
In this case we have 286 days from the beginning of the year to the 12th 
of October. 


♦This being an English date it is of course counted O. S„ as it hap¬ 
pened prior to 1752 










62 


THE SHORT-RULE ARITHMETIC. 


NEW STY1E. 

In devising a rule for telling the day of the week of any 
date according to New Style—which is our present method 
of reckoning time—we will first present a rule for the present 
century which is quite easy, and as there are no numbers, to 
remember it should be known by every one, as it will i iot 
only answer as a calendar for daily use when no printed 
calendar is at hand, but all dates past and future can ne 
easily ascertained. Corrections can be made from it for any 
century. 

* RULE.—To the given year add its one-fourth part 
(discarding fractions, if any)—in leap year add its one- 
fourth part less one —and the number of days that have 
elapsed since the beginning of the given year , inclusive. 
Divide their sum by 7. The excess of sevens signifies the 
day of the week. 1 signifies Sunday , 2 Monday , etc. An 
excess of 0 indicates Saturday. 

The battle of New Orleans was fought Jan. 8,1815; find 
the day of the week. 


Operation. 

1815 = Years. 

4 5 3 = Leap years. 

8 = Days elapsed since beginning of year. 
7)2276 

3 2 5 — 1= Sunday. 


Find the day of the week of July 4,1876. 


* In leap year do not forget the 29th of February when adding the 
number of days that have elapsed since the beginning of tne year. Ex¬ 
cept in the case of the centurial years before mentioned leap year can 
always be told after dividing the year by 4, when the quotient is a whole 
number. Remember to deduct one in such years. 






CHRONOLOGY. 


63 


Operation. 

18 7 6 = Years. 

4 6 8 = Leap years less one . 

31 = Days in January. 

2 9 = Days in February. 

31 = Days in March. 

3 0 = Days in April. 

31 = Days in May. 

3 0 = Days in June. 

4 = Days in July. 

7)2530 

3 61 — 3 = Tuesday. 

Explanation.— If the student will remember that for dates in the 
past century add 2, and for the next century add 5 to the operation be¬ 
fore dividing by 7 the dates for 300 years can be easily ascertained. 
Thus, suppose we wish to find on what day of the week occurred the 
signing of the declaration of independence July 4, 1776. We proceed as 
above, with the date in 1876 and add 2 to the sum, which gives 2532. 
2532-(-7=361, and a remainder of 5, which signifies Thursday, answer. In 
like manner, for the same date in 1976 we add 5 and get 2535, which 
divided by 7, as before, gives an exceess of 1, which signifies that July 
4, 1976, will fall on Sunday. 

The rule for telling the number of days to add for any century, New 
Style, is as follows: Find the figure to add for dates between 2000 and 
the century following. We reject the first two ciphers and we have 20. 
From this subtract 16 and multiply the remainder (4) by 51 (rejecting 
fractions) we get 21. We now add 4 to this result, which gives 25. 
Divide by 7, and we have 3 and a remainder of 4. The remainder is the 
figure sought. 

A rule which applies to the present century and by mak¬ 
ing corrections as in the preceding rule to all centuries— 
New Style—can be applied mentally. To do this, figures 
which we shall term “excess figures” of the months must be 
memorized. They are as follows: 


0 

1 

2 

3 

4 

5 

6 

June. 

Sept. 

Dec. 

April 

July 

Jan.* 

Oct. 

May 

Aug. 

Feb* 

Nov. 

March. 


* January and February are each called one less in leap years. 















64 


THE SHORT-RULE ARITHMETIC. 


Find the day of the week of Dec. 25,1858. 

8)580 = Last two figures of year with cipher annexed. 

7 2 = 1 Discarding fraction. 

2 5 = Date of month. 

1 = Excess figure of December. 

7) 98 

14 — 0 = Saturday. 

Great rapidity can be attained by rejecting the sevens as they occur. 
For instance, after dividing by 8 we have 72. 7 is contained in 72 10 
times and a remainder of 2. Simply retain the remainder. Adding this 
to 25 we get 27, and discarding the sevens we have 6, which added to the 
excess figure (1) of December makes 7. As this is an even number of 
sevens we call the date Saturday, as previously explained. 


MOON’S AGE. 


Table of Monthly Numbers. 


January 0 
February 2 
March 1 
April 2 


May 3 
June 4 
July 5 
August 6 


September 8 
October 8 
November 10 
December 10 


Find the age of the moon on the 20th of March, 1876. 


Explanation.— To the given year we add 1 which gives 1877, and di¬ 
vide by 19, we have 98 and a remainder of 15. Retain the remainder un¬ 
less it happens to be 0, when we must call it 19. We now multiply our 
remainder (15) by 11 which gives 165. From this substract the number of 
days suppressed by the reformation of the Calendar, which is 10 days if 
the year is between 1582 and 1700, 11 days if between 1700 and 1900,12 
days if between 1900 and 2200. According to this the number of days to 
substract from 165 is 11, which gives 154. We now divide 154 by 30, 
which gives 5 and a remainder of 4, which (4) added to the date of the 
month (20) we have 24. We also add the monthly number (1) which is 
found in the table and we have 25, which is the age of the moon on 
March 20 in 1876. If this last remainder be 30 it shows that the new 
moon took place on that day; but if it exceeds 30 substract 30 from it 






CHRONOLOGY. 


65 


and the remainder will be the age of the moon. In calculations of this 
nature great exactnesss must not be expected. The irregular arrange¬ 
ments of the months, the mean numbers necessary to be assumed in the 
formation of the periods from which these calculations are deduced 
and the irregularity of the lunar revolution may occasion an error of 
nearly 48 hours. 


EASTER SUNDAY. 


Easter Sunday should be celebrated on the first Sunday after the 14th 
day of the moon, if this 14th should happen on or after the 21st of 
March; therefore, Easter cannot happen earlier than March 22 nor later 
than April 25. The following table has been computed for easy reference: 


EASTER TABLE. 

Showing Easter Sunday for a period of 100 years, from which can be 
reckoned the dates of the celebration of Mardi Gras and all Church 
Feasts depending on Easter. 

Septuagesima Sunday is 9 weeks, First Sunday in Lent is 6 weeks, Ash 
Wednesday is 46 days, and Mardi Gras is 47 days before Easter. 

Rogation Sunday is 5 weeks, Ascension Day or Holy Thursday is 39 
days, Pentecost or Whit Sunday is 7 weeks, and Trinity Sunday is 8 
weeks after Easter. 


1800.. . April 13 

1801.. .April 5 

1802.. . April 18 

1803.. . April 10 

1804.. . April 1 

1805.. . April 14 

1806.. . April 6 

1807.. . March 29 

1808.. . April 17 
1809. ..April 2 

1810.. .April 22 
1811. ..April 14 

1812.. .April 29 

1813.. . April 18 

1814.. . April 10 

1815.. . March 26 

1816.. . April 14 

1817.. . April 6 

1818.. . March 22 

1819.. . April 11 

1820.. . April 2 

1821.. .April 22 

1822.. . April 7 

1823.. . March 30 

1824.. . April 18 

1825.. . April 3 


1826.. .March 26 

1827.. . April 15 

1828.. . April 6 

1829.. . Aoril 19 

1830.. . April 11 

1831.. . April 3 

1832.. . April 22 

1833.. . April 7 

1834.. . March 30 

1835.. . April 19 

1836.. . April 3 

1837.. .March 26 

1838.. . April 15 

1839.. .March 81 

1840.. . April 19 

1841.. . April 11 

1842.. . March 27 

1843.. . April 16 

1844.. . April 7 

1845.. . March 23 

1846.. . April 12 

1847.. . April 4 

1848.. .April 23 
1849. ..April 8 

1850.. . March 31 

1851.. . April 20 


I 1852... April 11 

1853.. . March 27 

1854.. . April 16 

1855.. . April 8 

1856.. . March 23 

1857.. . April 12 

1858.. . April 4 

1859.. .April 24 

1860.. . April 8 

1861.. . March 31 

1862.. .April 20 

1863.. . April 5 

1864.. . March 27 

1865.. . April 16 

1866.. . April 1 

1867.. .April 21 

1868.. . April 12 

1869.. . March 28 

1870.. . April 17 

1871.. . April 9 

1872.. . March 31 

1873.. . April 13 
1874. ..April 5 

1875.. . March 28 
1876. ..April 16 

1877.. . April 1 


1878.. .April 21 

1879.. . April 13 

1880.. . March 28 

1881.. . April 17 

1882.. . April 9 

1883.. . March 25 

1884.. . April 13 

1885.. . April 5 

1886.. .April 25 

1887.. . April 10 

1888.. . April 1 

1889.. .April 21 
1890. ..April 6 

1891.. . March 29 

1892.. . April 17 
1893. ..April 2 

1894.. . March 25 

1895.. . April 14 
1896. ..April 5 

1897.. . April 18 

1898.. . April 10 
1899. ..April 2 

1900.. . April 15 
1901. ..April 7 

1902.. . March 30 

1903.. . April 17 


5 











66 


THE SHORT-RULE ARITHMETIC. 


Mei^sUratIoiV 

-OB- 

Practical Geoipetry. 


These pages present in simple form and language a ready 
reference for civil engineers, mechanics, farmers, teachers, 
students, and all who have occasion to refer to this impor¬ 
tant branch of mathematics. Many of the rules are entirely 
original, and the entire collection cannot fail to meet all 
requirements in measuring surfaces and solids that are met 
with in practical work. 

SQUARES AND RECTANGLES. 


GIVEN TO FIND 

length and Breadth. Area of any Square or Rec¬ 

tangle. 

Rule.— Multiply the length by the hreadtli . 

Problem.—A field 10 rods long and 5 rods wide contains how many 
square rods? 

Operation.— 10X5=50 sq. rods, area. 

Remark.—Area, of a square = one-half the square of its diagonal. 


GIVEN 


TO FIND 


Area and one side of a Square 
or Rectangle. 


The Unknown Side of any 
Square or Rectangle. 


Rule.— Divide the area by the given side. 


A field 10 rods long and containing 50 square rods, what is the width. 
Operation— 50+10=5rods, width. 









MENSURATION. 


67 


GIVEY TO FI YD 

Side of a Square. Diagonal of a given Square. 

Bulb.—M ultiply the length of the given Side by 1.41421; 
or, multiply the length of the given side by 99 and divide 
the product by 70. 

The side of a given square is 4 feet. Find the diagonal. 

Operation.— 1.41421X4 (length of side) =5.65684, or 99 X 4= 396. 396-S-70 
=5.657, answer. 

Remark. The second rule gives slightly too much, but the error is of 
little consequence. 

GIVEY I TO FIYD 

Diagonal of a Square. | The Side of a given Square. 

Rule .—Multiply the diagonal of the square by .707107; or, 
multiply the diagonal by 70 and divide by 99. 

The diagonal of a given square is 7 feet. Find the length of a side. 

Operation. —.707107 X7 (diagonal)—4.949749 feet, length of side. 

Remark— When figures are in the form of a rhombus, rhomboid, 
trapezium, or trapezoid care should be used in ascertaining the true 
width. Frequently a side of such a figure is mistaken for the width. 


GIVEY ] TO FI YD 

Areaor Diagonal of a Square Area of the Circumscribed 

Circle of a given Square. 

Rule. —Double the area and multiply the product by 
.7854; or, multiply the square of the diagonal by .7854. 

The area of a given square is 8 feet. Find the area of its circum¬ 
scribed circle. 

Operation— 8(area)x2=16. .7854X16=12.5664sq. ft., answer. 

Diagonal of a given square is 4 feet. Find the area of its ciroum- 
scribed. 

Operation.—& (diagonal) =16. .7854X16=12.5664 sq. ft., answer. 
Remark— It should be remembered that the area of an inscribed cir¬ 
cle of a square is one-half the area of its circumscribed circle, and the 
same proportion exists between inscribed and described squares of cir¬ 
cles. 







68 


THE SHORT-RULE ARITHMETIC. 


GIVEN TO FIND 

Area or Diagonal of a Square Area of the Inscribed Circle 

of a given Square. 

Rule.— Multiply the area by .7854, or take one-half the 
square of the diagonal and multiply by .7854. 

The area of a given square is 8 feet. Find the area of its inscribed 
circle. 

Operation.—. 7854X8=6.2832 sq. ft. area of inscribed circle. 

Diagonal of a given square is 4 feet. Find the area of its inscribed 
circle. 

Operation .—42 (diagonal) =16. 16-*-2=8. .7854X8=6.2832 sq. ft. area of 
inscribed circle. 


GIVEN 


TO FIND 


One Side or Diagonal of a 
Square. 


Radius of the Circumscribed 
Circle of a Square. 


Rule.— Multiply the given side by .707107; or, take one- 
half the diagonal. 


The side of a given square is 5 feet. Find the radius of its circum¬ 
scribed circle. 

Operation.— .707107X5(side) =3.535535 ft., answer. 


GIVEN 

Side of a Square. 


TO FIND. 

Diameter of a Circle of Equal 
Area. 


Rule.— Multiply the given side by 1.12838. 


The side of a given square is 3 feet. Find the diameter of a circle con 
taining an equal area. 

Operation.— 1.12838 X 3 (length of side) =3.38514 ft., answer. 


GIVEN TO FIND 

Side of a Square. The Side of an Equilateral 

Triangle of Equal Area. 

Rule.— Multiply the given side by 1.51967. 

The side of a given square is 3 feet. Find the side of an equilateral 
triangle which will contain an equal area. 

Operation.— 1.51967X3 (side of given square) =4.55901 ft., answer. 










MENSURATION. 


69 


GIVEN 

Length of One Side of a Reg¬ 
ular Polygon. 


TO FIND 

Area of a Regular Polygon. 


Rule.— Square one side of the polygon and for the 


Equilateral triangle (3 sides) X 
Pentagon (5 sides) X 


Hexagon (6 sides) X 

Heptagon (7 sides) X 

Octagon (8 sides) X 

Nonagon (9 sides) X 

Decagon (10 sides) X 

Undecagon (11 sides) X 

Dodecagon (12 sides) X 11.196152 

Or, divide the polygon into triangles and find the sum of 
their areas ; or, multiply one-half the perimeter by the 
apothem. 

TRIANGLES. 


.433013 

1.720477 

2.598076 

3.633912 

4.828427 

6.181824 

7.694209 

9.3656*0 


GIVEN TO FIND 

Base and Altitude, or Three Area of any Triangle. 
Sides, or the Sum of Sides. 

Rule. —Multiply base by half the altitude; or, from the 
half sum of the three sides substract each sum separately, 
multiply the half sum and the three remainders continu¬ 
ously together and extract the square root of the product. 

Sides of a triangle are 4, 7, and 9 feet, find area. 

Operation .—*4 + 7 + 9=20, sum of sides. 20+2=10, half sum of sides. 
10-4=6. 10-7=3. 10-9=1. 10X6X3X1=180. Square root of 180= 
13.4 + square feet area. 


GIVEN. TO FIND 

Area and Altitude of a Tri- The Base of any Triangle, 
angle. 

Rule. —Divide twice the area by the altitude. 

The area of a given triangle is 6 feet, altitude 3 feet. Find base. 
Operation— 6(area)X3=12. 12+3(altitude)=4 ft., length of base. 









70 


THE SHORT-RTJLE ARITHMETIC. 


GIVEX TO FIND 

Area and Base of a Triangle. Altitude of any Triangle. 

Rule Divide twice the area by the base. 

The area of a given triangle is 6 feet, base 4 feet. Find altitude. 
Operation .—6 (area)X2=12. 12-*-4 (base)=3ft., altitude. 


GIVEN I TO FIND 

The Three Sides of any Tri- I Diameter of the Circnm- 
angle or their product. scribed Circle of any Tri- 
| angle. 

Rule.—D ivide the product of the sides of the triangle by 
twice the area. 

The sides of a given triangle are 3, 4, and 5 feet. Find the diameter 
of its circumscribed circle. 

Operation— 3X4X5=60, product of the sides. The area, found by a 
preceding rule is 6 ft. Twice the area =12 ft. 60-s-12=5 ft., diameter of 
circumscribed circle. 


GIVEN 


TO FIND 


Side of an Equilateral Tri¬ 
angle. 


Area of an Equilateral Tri¬ 
angle. 


Rule.— Multiply the square of the given side by .433013; 
or, multiply the apothem by half the perimeter (sum of 
sides). 

One side of an equilateral triangle is 4 feet. Find the area. 
Operation— 42 (side)=16. .433013X16 =6.928208 sq. ft., area. 


GIVEN 


TO FIND 


Side of an Equilateral Tri¬ 
angle. 


Altitude of an Equilateral 
Triangle. 


Rule.— Multiply the given side by .866025. 

The side of a given equilateral triangle is 4 feet. Find the altitude. 
Operation—. 866025X4 (side) =3.464100 ft., altitude. 











MENSURATION. 


71 


OIVEX | TO FIXD 

Altitude of an Equilateral I Side of an Equilateral tri¬ 
triangle. | angle. 

Rule. —Multiply the altitude by 1.1547, or divide the al¬ 
titude by .866025, or multiply the square root of the area 
by 1.51967. _ . 


(ilVEX 


TO FIND 


Side of an Equilateral Tri¬ 
angle. 


Side of a Square of equal 
area. 


Rule.— Multiply the given side by .658037. 


The side of a given equilateral triangle is 6 feet. Find the length of 
one side of a square which will contain the same area. 

Operation— .658037 X6 =3.948222ft., answer. 


GIVEX 

Side of an Equilateral Tri¬ 
angle. 


TO FIND 

Diameter of a Circle of 
equal area. 


Rule.— Multiply the given side of the triangle by .742517. 


What is the diameter of a circle which will contain the same area as 
an equilateral triangle whose sides Eire 6 feet each. 

Operation. —,742517X6(side) =4.455102 ft., diameter. 


GIVEX 

Side of an Equilateral Tri¬ 
angle. 


TO FIXD 

Diameter of the Circum¬ 
scribed Circle of an Equilat¬ 
eral Triangle. 


Rule.— Multiply the length of the given side by 1.1547. 

The side of an equilateral triangle is 4 feet. Find the diameter of its 
circumscribed circle. 

Operation— 1.1547 X 4=4.6188ft. diameter. 


GIVEN 


TO FIND 


Altitude of an Equilateral 
Triangle. 


Apothem or Diameter of the 
Inscribed Circle of an 
Equilateral Triangle# 


Rule#—T ake one-third of the altitude, 












72 


THE SHORT-RXJLE ARITHMETIC. 


Altitude of an equilateral triangle is 6 feet. Find the diameter of its 
inscribed circle. 

Operation .—6 (diameter) -4-3=2 ft., answer. 

Remark—One-fourth of the diameter of its circumscribed circle of an 
equilateral triangle also equals the diameter of its inscribed circle or 
apothem. 


GIVEN 


TO FIXD 


Base and Altitude of a Right 
angled Triangle. 


Hypotenuse of a Right An¬ 
gled Triangle. 


Rule.— Extract the square root of sum of the squares of 
the base and altitude. 


Base and altitude of a given right angled triangle are 5 feet and 12 feet 
respectively. Find the hyptenuse. 

Operation —52 +122=169. Square root of 169=13 ft., hypotenuse. 


GIVEN 


TO FIND 


Base and Hypotenuse of a 
Right Angled Triangle. 


Altitude of a Right Angled 
Triangle. 


Rule.— From the square of the hypotenuse substract the 
square of the base and extract the square root of the re¬ 
mainder. 

The hypotenuse and base of a right angled triangle are 13 feet and 5 
feet, respectively. Find the altitude. 

Operation.— 132=169. 52= 25. 169—25=144. Square root of 144=12 ft., 
altitude. 


GIVEN 

Altitude and Hypotenuse of 
a Right AngledgTriangle. 


TO FIND 

Base of a Right Angled Tri¬ 
angle. 


Rule.— From the square of the hypotenuse substract the 
square of the altitude and extract the square root of the re¬ 
mainder. 


The hypotenuse and altitude of a right angled triangle are 13 feet and 
12 feet respectively. Find the base. 

Operation— 132=169. 122=144, 169-144=25. Square root of 25=5 ft., 

base. 










MENSURATION. 


73 


THE CIRCEE. 


GIVEN TO FIND 

Diameter or Area of a Circle. Circumference of a Circle. 

Rule. —Multiply the diameter by 3i, or 3.1416, or for 
greater “accuracy” by 3.141592653, or multiply the area by 
12.566 and extract the square root of the product. 

Remark .—Archimedes found that the circumference of any circle was 
nearly 8} times the diameter. If the exact ratio could be found the 
process of squaring or quadraturing the circle would be accomplished 
by figures. Thousands of persons have claimed in almost as many differ¬ 
ent ways the accomplishment of this impossibility. The ratio of the 
diameter to the circumference has been found to over 600 decimal places. 
The preceding ratios will answer all practical purposes, but we will ap¬ 
pend the following, which gives the ratio of the diameter to the circum¬ 
ference to be as 

1 to 8.141592653589793238462648383279502884197169399375105820974944592807816 
4062862089986280348253421170679821480865132823066470938 + to infinity. 


GIVEN TO FIND. 

Circumference or Area of a Diameter of a Circle. 
Circle. 

Rule.— Multiply the circumference by .31831, or multiply 
the area by 1.2732 and extract the square root of the pro¬ 
duct. 

The circumference of a given circle is 9 feet. Find the diameter. 
Operation.— .31831X9 (circumference! =2.86479 ft., diameter. 


GIVEN. TO FIND 

Diameter or Circumference Area of a given Circle, 
of a Circle. 

Rule. —Square the diameter and multiply by .7854, or 
for greater “accuracy” multiply by .78539816, or square the 
circumference and multiply by .07957747, or multiply half 
the diameter by half the circumference, or square the ra¬ 
dius and multiply by 3.1416. 










74 


THE SHORT-RXJLE ARITHMETIC. 


GIVEX 


TO FIXD 


Diameter or Circumference 
of a Circle. 


Side of the Inscribed Square 
of a Circle. 


Rule.— -Multiply the diameter by .707107, or multiply the 
circumference by .22508. 


The diameter of a given circle is 8 feet. Find the side of the largest 
square that can be inscribed therein. 

Operation— .707107X8(diameter) =5.656856ft., answer. 


GIVEX TO FIXD 

Diameter of a Circle. Area of the Inscribed Square 

of a Circle. 

Rule.— Take one-half the square of the diameter. 


GIVEN 


TO FIND 


Diameter of Circle. 


Areaof the Described Square 
of a Circle. 


Rule.— Square the diameter. 


GIVEN TO FIXD 

Diameter of a Circle. Side of an Equilateral Tri¬ 

angle of equal Area. 

Rule. —Multiply the diameter by 1.8468. 


CIRCULAR ARCS A AD CHORDS. 


GIVEN 


TO FIND 


JLength of Arc and Circum¬ 
ference of a Circle. 


Number of Degrees in a Cir¬ 
cular Arc. 


Rule.— Multiply the length of the given arc by 360 and 

divide by the circumference . 














MENSURATION. 


75 


The circumference of a given circle is 50 feet, and the length of the 
given arc is 8 feet. Find the number of degrees in the arc. 

Operation— 360X8=2880. 2880+ 50= 57g, answer. 


GIVEX TO FIXD 

The Degrees of an Arc of a ^Length of a Circular Arc. 
Circle. 

Rule. —Multiply the number of degrees by the circum¬ 
ference and divide by 360, or multiply the chord of half the 
arc by 8, substract the chord of the whole arc and divide 
the remainder by 3, or multiply the number of degrees in 
the arc by the diameter of the circle and the product by 
.0087266. 


GIVEN | TO FIND 

Height of Arc and Chord of I Diameter of the Circle of a 
half the Arc. I given Arc. 

Rule.— Divide the square of the chord of half the arc by 
the height of the arc. 

GIVEN TO FIND 

Chord of half the A rc and Di- Height of a Circular Arc. 
ameter of the Circle. 

Rule .—Divide the square of the chord of half the arc by 
the diameter of the circle. 

The chord of half an arc is 12 feet and the diameter of the circle is 36 
feet. Find the height of the arc. 

Operation.— 122=144. 144X36=4 ft., height of arc. 

GIVEN | TO FIND 

Height of Arc and Diameter I The Chord of Half the Arc. 
of the Circle. I 

Rule. —Multiply the diameter of the circle by the height 
of the arc and extract the square root of the product. 

The height of an arc is 4 feet, and the diameter of the circle is 36 feet. 
Find the length of the chord of half the arc. 

Operation .—36 X 4=144. Square root of 144=12 ft., length of chord of 
half the arc. 









76 


THE SHORT-RTJLE ARITHMETIC. 


GIVEN j TO FIND 

Length of Chord and Height Diameter of a Circle of a Cir- 
of an Arc. I cular Arc. 

Rule .—Divide the square of half the chord by the height 
and to the quotient add the given height. 

The length of a given chord is 56 feet and the height 8 feet. Find the 
diameter of the circle. 

Operation .—56 (length of chord) -4-2=28. 28 2 =784. 784+8 (height)=98. 
98+8=106 ft., diameter of the circle. 


GIVEN | TO FIND 

The Height of an Arc and Di- I Length of the Chord of a Cir- 
ameter of the Circle. | cular Arc. 

Rule. —Multiply the difference of the height of the arc 
and the diameter of the circle by the height of the arc ; 
extract the square root of the result and multiply by 2. 

The height of an arc is 2 feet and the diameter of the circle is 10 feet. 
Find the length of the chord of the arc. 

Operation —10 (diameter)—2 (heightof arc)=8. 8X2 (height of arc)= 
16. Square root of 16=4. 4X2=8 ft., chord of arc. 


GIVEN TO FIND 

The Length of the Chord of an Height of a Circular Arc. 

Arc and Diameter of the 

Circle. 

Rule. —Take half the diameter of the circle and half the 
Tenath of the chord of the arc. Multiply their sum by their 
difference, extract the square root of the product, and sub¬ 
tract the square root thus found from half the diameter of 
the circle. 

The chord of an arc is 8 feet and the diameter of the circle is 10 feet. 
Find the height of the arc. 

Operation.— 4=4 diameter. 5=1 length of arc. 4 + 5=9, sum of half 
the diameter and half the length of the arc. 5—4=difference of half the 
diameter and half the length of the arc. 9X1=product of difference 
and sum. Square root of 9=3. 5 (half diameter-3 (sq. xt.)=2 ft., 
height of arc. 







MENSURATION. 


77 


\ 


GIVEX 


TO FIND 


Length of Chord and Diam¬ 
eter of the Circle. 


Chord of Half a Circular 
Arc. 


Rule.— Find the height of the arc by the preceding rule, 
and to its square add the square of half the length of the 
chord of the arc and extract the square root of the sum. 

The chord of an arc is 14 feet and the diameter of the circle is 50 feet. 
Find the length of the chord of half the given arc. 

Operation .—The height of the arc, found by the preceding rule, is 1 ft. 
Half of the chord (14 ft.) =7. 72=49, to which add the square of the 
height (12) 1, we get 50. Square root of 50=7.071 +ft., length of cord of 
half the given arc. 


GIVEN TO FIND 

Chord of Half an Arc and Di- The Chord of a Circular Arc. 
ameter of the Circle. 

Rule.— Divide the square of the chord of half the arc by 
the diameter of the circle , subtract the square of the quo¬ 
tient from the square of the given length of the chord of 
half the arc , and extract the square root. Then multiply 
the root thus found by 2. 

The chord of half an arc is 12 feet and the diameter of the circle is 36 
feet. Find the chord of the given arc. 

Operation —122 (length of chord of half the arc) =144. 144-5-36 (diam¬ 
eter of circle) =4. 42=16. 122 (chord of half the arc)=144. 144-16=128. 
Square root of 128=11.31 + . 2Xll.31=22.62+ft., length of chord of arc. 


GIVEN 


TO FIND 


Chord of an Arc and Chord of 
Half the Arc. 


The Diameter of the Circle 
of a Circular Arc. 


Rule.— From the square of the chord of half the arc sub¬ 
tract the square of half the chord of the whole arc and ex¬ 
tract the square root of the remainder. Divide the square 
of the chord of half the arc by the root thus found. 

The chord of an arc is 48 feet and the chord of half the arc is 26 feet. 
Find the diameter of the circle. 

Operation .—48 (chord of whole arc)+2=24. 242= 576. 26 2 (chord of 
half the arc)=676. 676 -576=100. Square root of 100=10. 676 (square of 
chord of half the arc) -*-10=67.6 ft., diameter of circle. 








78 


THE SHORT-RULE ARITHMETIC. 


GIVEK TO FIXD 

Length of Arc and Radius of Area of a Sector of a Circle. 

Circle, or Area of Circle 

and Degrees of Angle of 

Sector. 

Rule. —Multiply the length of the arc by the radius and 
take half the product; or, multiply the area of the circle 
by the number of degrees in the angle of the sector and 
divide by 360. 

Problem. —The radius of a circle is 10 feet and the angle of the sec¬ 
tor 40 degrees. Find the area of the sector. 

Operation .—First find area of circle. 10 (radius) X2=20 diameter. 202= 
400. .7854 X400= 314.1600 sq. ft., area of circle. 314.16 (area) X40 (No. de¬ 
grees in angle of sector) =12566.40, which divided by 360=34.90+sq. ft., 
area of sector. 

Problem 2.— (Length of arc given.) The length of the arc is 6 feet 
and the radius 8 feet. Find the area of the sector. 

Operation.— 8X6=48. 48+2=24 sq. ft., area of sector. 


GIVEX TO FIXD 

Angle and Radius of a Circle. Area of a Segment of a 

Circle. 

Rule. —Find the area of the sector which has the same 
arc and substract the area of the triangle formed by the 
radii and the chord. 

The radius of a circle is 10 feet, and the angle of the sector is 60 de¬ 
grees. Find the area of the segment. 

Operation— 10X10X3.1416=314.16, area of circle. As the area of the 
sector is 60 degrees or | of the circle, its area is | of 314.16 or 52.36 sq. ft. 
The triangle in this case is equilateral and the area 43.30ft. subtracted 
from 52.36 = 9.06 sq. ft. area of segment. 


GIVEN 


TO FIND 


Iiengtli of Chord and Height 
of Arc of a Circle. 


Area of a Segment of a 
Circle. 


Rule. —Add together one-fourth of the square of the 
chord and two-fifths of the square of the height and multi¬ 
ply the square root of the sum by | of the height. 









MENSURATION. 


79 


This rule taken from Todhunter’s Mensuration is not exact, as it gives 
the area of the segment greater than it ought to be, but the error is very 
small provided the angle of the corresponding sector is small; when this 
area is 60 degrees, the error is less than part of the area, and when 
this angle is 90 degrees the area is less than part of the area. 


THE ELLIPSE. 


GIVEN TO FIND 

Uotli Diameters. Area. 

Rule.— Multiply the product of both diameters by .7854. 


RECTANGULAR SOLIDS. 


GIVEN! TO FIND 

Length, Width, and Height. Cubic Contents of a Cube or 

any Rectangular Solid. 

Rule.— Multiply together the length , width , and height. 

A piece of timber 10 inches long, 6 inches high, and 4 inches wide. 
Find the cubic contents. 

Operation.— 10X6X4=240 cu. in., solidity. 

Remark .—To change cubic inches to cubic feet divide cubic inches 
by 1728. _ 


GIVEX 


TO FIXD 


Cubic Contents and Lengths 
of Two Sides. 


Length of an Unknown Side 
of a Rectangular Solid. 


Rule.— Divide the cubic contents by the product of the 
tiuo given sides . 

What should be the length of a box necessary to contain 240 cubic 
inches whose height is 6 inches and width 4 inches? 

Operation.— 4 (width) X6 (height)=24. 240+24=10 in., length. 


GIVEN 


TO FIXD 


Lengt h of a Side of a Cube. 


Diagonal of a Cube or Diam¬ 
eter of its Circumscribed 
Sphere. 


Rule.— Multiply the length of the side by 1.7320508. 














80 


THE SHORT-RULE ARITHMETIC .. 


PRISMS. 


GITEX TO FOD 

length of Side and Height of Cubic Contents of a Prism, 
a Prism. 

Rule.— Multiply the area of the base by the height. 

Remark .—The base of a prism is usually one of the regular polygons 
The rules for finding their areas are on page 69. 

Find cubic contents of a triangular prism, the length of one side of 
the base being 4 feet and the height 8 feet. 

Operation— We find the area of the base by rule on page 69. Thus, 
42=16. .433013X16=6.928208, sq. ft. area of base, which multiplied by 8 
(height) given 55.425664 cu. ft., answer. 


ttlVEN TO FIKD 

Cubic Contents and Area of The Height of a Prism, 
the Base of a Prism. 

Rule.— Divide the cubic contents by the area of the base. 

Find the height of a hexagonal prism necessary to contain 40 cubic 
feet. The length of each side is 9 feet. 

Operation .—We find the area of the base, as in the preceding example, 
to be 23.31684 sq. ft., which divided by 40 (desired contents)=1.7154 ft. 
high, answer. 

Remark .—The decimal of a linear foot can be reduced to inches by 
multiplying it by 12. Thus, .7154X12=8.5848 linear inches, i. e., a little 
more than 81 inches. 


GIVElff 


TO FIXD 


Perimeter and Height of a 
Prism. 


Area of the lateral Surface 
of a Prism. 


Rule. —Multiply the perimeter by the height. 

The length of a side of a pentagonal prism (5 sided) is 6 inches and 
its height 20 inches. Find the area of its lateral surface. 

Operation .—5 (number of sides) X6 (length of one side) =30 (perimeter, 
or stun of all sides). 30 X 20=600 sq. in., area. 

Remark .—Lateral surface means surface of the sides. If the area of 
the whole surface is required, add to the lateral surface the areas of the 
ends. 










MENSURATION. 


81 


GIVEN TO FIND 

Area of the Base and Height Cubic Contents of a Prism, 
of a Prism. 

Rule.—M ultiply the area of the base by the height and 
take one-third of the product. 

The side of a quadrangular prism (4 sided) is 6 feet and the height 10 
feet. Find the cubic contents. 

Operation.— 2=36 sq. ft., area of base. 36X10 (heightj =360. 360+ 3= 
120 cu. ft., contents. 

Remark— It will thus be seen that the contents of any pyramid is 
one-third of the contents of a prism of equal base and height. 


PYRAMIDS. 


GIVEN 


TO FIXI> 


The Slant Height and Perim¬ 
eter of a Pyramid. 


The Area of the lateral Sur¬ 
face of a Pyramid. 


Rule.—M ultiply the perimeter of the base by the slant 
height and take one-half the product. 


Find the lateral surface of a triangular pyramid the perimeter of 
which is 27 inches and slant height 18 inches. 

Operation— 18(slantheight)X27 (perimeter)=486. 486 +2=243 sq. in., 
area of lateral surface. 

Remark— It should be known that pyramids, as well as cones, have 
both a true and slant height. The true height is the shortest distance 
from the apex to the base. The slant height is the shortest distance from 
the apex to the perimeter. 


GIVEN TO FIND 

Area of the Base and Cubic The Height of a Pyramid. 
Contents of a Pyramid. 

Rule.—D ivide three times the contents by the area of the 
base. 

Remark— This rule also applies to cones. 

The area of the base of a pyramid is 40 square inches and the contents 
is 300 cubic inches. Find the height. 

Operation .—300 (cu. contents) X3 =900. 900+40 (area of base) =221 in. 
high. 









THE SHORT-RULE ARITHMETIC. 


GIVEN 


TO FIND 


Cubic Contents and Height 
of a Pyramid. 


The Area of the Base of a 
Pyramid. 


Rule. —Divide the contents by the height. 


GIVEN 


TO FIND 


Areas of the Two Ends and The Cubic Contents of a 
Height of a Frustum of a Frustum of a Pyramid. 
Pyramid. 


Rule. —To the sums of the areas of the tivo ends of the 
frustum add the square root of their product. Multiply 
the sum by the height and take one-third of the product. 


Find the cubic contents of a frustum of a pyramid whose height is 15 
feet. The area of one end is 18 square feet and the other 98 square feet. 

Operation .—18 + 98=116 (area of the two ends). 98X18=1764 square 
root of 1764= 42. 116 + 42=158. 15 (height) X 158 =2370, which divided by 
3 gives 790 cu. ft., answer. 

Remark .—This rule also applies to frustums of cones, but further on 
we give a rule which is much shorter for such cases. 


GIVEN 


TO FIND 


Botb Perimeters of a Frus¬ 
tum of a Pyramid. 


The Area of the Surface of a 
Frustum of a Pyramid. 


Rule.— Multiply half the sum of the upper and lower 
perimeter by the slant height of the frustum. 

Each of the sides of a frustum of a hexagonal pyramid (6 sided) is 2 
feet at top and 4 feet at the bottom. The height is 12 feet. Find the 
area of the surface. 

Operation .—2 (width of one side at top)X6 (number of sides) =12, 
perimeter at top. 4 (width of one side at bottom)X6 (number of sides) = 
24, perimeter at bottom. 24 + 12=36. 36-^2=18, mean perimeter. 18X12= 
216 sq. ft., area. 

Remark .—This gives the area of the sides, to which add the areas of 
the ends if required. 


CYLINDERS. 


GIVEN 


TO FIXD 


Diameter and Height of a 
Cylinder. 


Cubic Contents of a Cylin¬ 
der. 


Rule.—M ultiply the square of the diameter by .7854 and 
the product by the height. 











MENSURATION. 


83 


Find the cubic contents of a cylinder 3 feet in diameter and 7 feet 
high. 

Operation— 32 (diameter) =9. .7854 X9= 7.0686, area of base. 7.0686 X7 
(height)=53.4802 cu. ft., answer. 


GIVEN 


TO FIND. 


Diameter and Height of a 
Cylinder. 


The Cylindrical Contents of 
a Cylinder. 


Rule.— Square the diameter and multiply by the height. 


GIVEN. TO FIND 

Diameter and Cylindrical Height of a Cylinder. 

Contents of a Cylinder. 

Rule.— Divide the cylindrical contents by the square of 
the diameter . 

Find the height of a cylindrical can necessary to contain 3718 cylin¬ 
drical inches. 

Operation— 132=169. 3718+169=22inches, height. 

Remark— Cylindrical contents multiplied by .7854=cubic contents. 


GIVEN TO FIND 

Height and Cylindrical Con- Diameter of a Cylinder, 
tents of a Cylinder. 

Rule.—D ivide the cylindrical contents by the height and 
extract the square root of the quotient . 

A cylindrical can 41 inches high and containing 8036 cylindrical 
inches. Find the diameter. 

Operation—SOW (cyl. contents)+41 (height)=196, circular area of 
base. Square root of 196=14 inches diameter. 


GIVEN 


TO FIND 


Circumference and Height 
of a Cylinder. 


The Curved Surface of a Cyl¬ 
inder. 


Rule. —Multiply the circumference of one of the ends by 
the height. 

Remark .—Add to the curved surface the areas of both ends if the area 
of the entire surface is required. 











84 


THE SHORT-RULE ARITHMETIC. 


COJfES. 


GIVEN 


TO FIND. 


Diameter of Base and True 
Height of a Cone. 


The Cubic Contents of a 
Cone. 


Rule.— Multiply the square of the diameter of the "base 
by .7854 and the product by the height and divide by 3. 


Remark. —It will be observed that the contents of a cone is one-third 
of a cylinder of equal base and altitude. 

The diameter of a cone at the base is 4 feet and the height 6 feet. Find 
cubic contents. 

Operation. —42=16. .7854X16=12.5664 sq. ft., area of base. 12.5664X6 
(height)=75.3984 cubic contents of a cylinder of equal base and altitude, 
which divide by 3=25.1328 cubic contents of cone. 


GIVEN 


TO FIND 


Diameter and Slant Height 
of a Cone. 


Area of the Curved Surface 
of a Cone. 


Rule.—M ultiply the circumference of the base by the 
slant height and divide by 2. 

Remark.— To which add the area of the base if the whole surface is re¬ 
quired. 

Find the curved surface of a cone whose slant height is 6 feet and 
diameter 2 feet. 

Operation.— We first find circumference of base. 3.1416X2(diameter) = 
6.2832, which we multiply by the slant height (6), which gives 37.6992. 
We take one-half of the result and we have 18.8496 sq. ft., surface. 

Remark 2.—If the diameter and true height are given we can find the 
slant height by squaring the true height and adding the product to the 
square of the radius and extract the square root of their sum. 


GIVEN 


TO FIND 


Diameter and Altitude of a 
Cone. 


Cylindrical Contents of a 
Cone. 


Rule. —Multiply the square of the diameter by the alti¬ 
tude and divide the product by 2. 









MENSURATION. 


85 


GIVEX TO FIXD 

Diameter and Altitude of a Cubic Contents of a Cone. 
Cone. 

Rule.— Find the cylindrical contents by the preceding 
rule and multiply by .7854. 


GIVEX TO FIND 

Cylindrical Contents and The Altitude of a Cone. 
Diameter of a Cone. 

Rule.— Multiply the cylindrical contents by 3 and divide 
the product by the square of the diameter . 

The diameter of a cone is 4 feet and the cylindrical contents is 48 feet. 
Find the altitude or true height. 

Operation— Cyl. ft.=48X3=144. 42=16 square of diameter, 144+16=9 
ft., altitude. 


GIVEN TO FIND 

Altitude and Cylindrical Diameter of a Cone. 

Contents of a Cone. 

Rule.— Multiply the cylindrical contents by 3, divide the 
product by the altitude and extract the square root of the 
quotient. 

The height of a cone is 12 feet and the cylindrical contents are 196 feet. 
Find the diameter. 

Operation.—196 X 3= 588. 588+12= 49. Square root of 49=7 ft., diam¬ 
eter. 


GIVEN. 

The Upper and Lower Diam¬ 
eters and Height of a Frus¬ 
tum of a Cone. 


TO FIND 

Cylindrical Contents of a 
Frustum of a Cone. 


Rule. —To the sum of the squares of the upper and lower 
diameters add the product of both diameters. Multiply the 
complete sum by the height and divide by 3. 

Find the cylindrical contents of a tank in the form of a frustum of a 
cone whose upper and lower diameters are 3 and 5 feet respectively, and 
whose height is 7 feet. 

Operation— 32=9 square of upper diameter. 52=25 square of lower 
diameter. 3X5=15 product of both diameters. 9 + 25 + 15=49. 7 (height) 
<49= 343. 343 + 3=114$ cyl. ft., contents. 











86 


THE SHORT-RULE ARITHMETIC. 


GIVEN 

Upper and Dower Diameters 
and Height of a Frustum of 
a Cone. 


TO FIND 

Cubic Contents of a Frus¬ 
tum of a Cone. 


Rule— Find the cylindrical contents by the preceding 
rule and multiply by .7854. 


GIVEN TO FIND 

The Two Diameters and The Height of a Frustum 
Cylindrical Contents of of a Cone, 

a Frustum of a Cone. 

Rule—D ivide the cylindrical contents by one-third of 
the product of the two diameters and their squares. 

A frustum of a cone whose upper diameter is 3 feet and lower diame¬ 
ter 6 feet contains 252 cylindrical feet. Find the height. 

Operation— Square of upper diameter=9. Square of lower diameter 
=36. 6X3=18, product of the two diameters. 9 + 36-5-18=63. 63-5-3=21. 
252 (cyl. ft.)-5-21=12 ft., height. 


GIVEN 


TO FIND 


The Two Diameters and Slant The Curved Surface of a Frus- 
Heiglit of a Frustum of a turn of a Cone. 

Cone. 

Rule.—M ultiply one-half the sum of the upper and lower 
circumferences by the slant height. 


THE GLOBE OR SPHERE. 


GIVEN TO FIND 

The Diameter of a Sphere. The Cubic Contents of a 

Sphere. 

Rule.— Cube the diameter and multiply by .5236. 

How many cubic feet are contained in a round ball or globe whose 
diameter is 5 feet? 

Operation— 5X5X5=125. .5236X125=65.45 cu. ft., volume. 

Remark .—The volume of a sphere also equals the cube of its circum¬ 
ference multiplied by .01689, or cube of radius multiplied by 4.1888, or 
its surface multiplied by one-sixth of diameter, or multiply the surface 
by diameter and divide by 6, or area of 4 circles of equal diameter mul¬ 
tiplied by one-sixth of its diameter. 












MENSURATION. 


87 


GITEX TO FIXD 

Diameter of a Sphere. Surface of a Sphere. 

Rule. —Square the diameter and multiply by 3.1416. 

Remark .—The surface of a sphere also equals square of circumference 
multiplied by .31831, or the diameter multiplied by the circumference, 
or four times the area of a circle of equal diameter, or the square of the 
radius multiplied by 12.5664, or the solidity divided by one-sixth of its 
diameter. 


GIVEX 


TO FIXD 


Inner and Outer Diameter of 
a Sperical Shell. 


Cubic Contents of a Spheri¬ 
cal Shell. 


Rule.— Subtract the cube of the inner diameter from the 
cube of the outer diameter and multiply the result by .5236. 

The inner diameter of a spherical shell is 7 inches and the outer diam¬ 
eter is 9 inches. Find the cubic contents of the shell. 

Operation.— The cube of 9 is 729. The cube of 7 is 343. 729-343—386. 
386X.5236 =202.1096 cu. in., contents. 


GIVEX 


TO FIXD 


Radii of the Two Ends and The Cubic Contents of a 
Height of a Spherical Zone. Spherical Zone. 


Rule.—A dd the square of the height to three times the 
sum of the squares of the radii of the two ends , and multi¬ 
ply the complete sum by the height and by .5236. 


The radii of the ends of a spherical zone are 8 and 10 inches and the 
height is 3 inches. Find the cubic contents. 

Operation .— 8 2 =64. 102=100. 64 + 100=164. 164X 3= 492. 492 + 9 (square 
of height)=501. 501X3 (heightJ= 1503. 1503X.5236= 786.9708 cu. in., con¬ 
tents. _ > 


GIVEX 


TO FIXD 


Height of Spherical Zone or 
or Segment and the Circum¬ 
ference of its Sphere. 


The Curved Surface of a 
Spherical Zone or Seg¬ 
ment. 


Rule.— Multiply the circumference of the sphere by the 
height of the zone or segment. 

The height of a segment or % sphere is 6 inches and the diameter of 
the sphere is 18 inches. Find tbd area of the curved surface. 

Operation .—We first find tne circumference, thus: 3.1416X18 (diame- 
ter)=56.5488. 56.5488X 6 (height) =339.2928 sq. in., surface. 











88 


THE SHORT-RULE ARITHMETIC 


GIVEN 


TO FIND 


Radius of the Base and Height 
of a Spherical Segment 


The Cubic Contents of a 
Spherical Segment. 


Rule.— To the square of the height add three times the 
square of the radius of the base. Multiply the sum by the 
height and by .5236. 


The radius of the base of a spherical segment is 5 inches and the 
height is 3 inches. Find the cubic contents. 

Operation .—52 (radius of base)=25. 25X3=75. 32=9 (square of height). 
75 + 9=84. 84X3 (height) =252. 252X.5236=131.9472cu. in. 










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LIBRARY of congress 



0 019 848 776 4 



H. W. FRIEDHOFF, 

GENERAL AGENT, 

62 Barton St., Cincinnati, O 


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